# 2x squared -5=4x use quadratic formula write the equation in standard form identify a,b,c solve by using quadratic formula & identify the solution set

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The equation 2x^2 - 5 = 4x has to be solved for x.

2x^2 - 5 = 4x

=> 2x^2 - 4x - 5 = 0

The standard form of the quadratic equation is ax^2 + bx + c = 0 and the solution of the equation is `(-b+-sqrt(b^2 - 4ac))/(2a)`

Here a = 2, b = -4 and c = -5

The roots of the equation are `(4+-sqrt(16 + 40))/(4)`

= `(4+-sqrt 56)/(4)`

= `(2*(2 +- sqrt 14))/4`

The roots of the equation are `(2 + sqrt 14)/2` and `(2 - sqrt 14)/2`

`2x^2 - 5 = 4x`

First, you need to recognize the general form of a quadratic function.

`ax^2 + bx + c`

Manipulate your given equation to match the one above by subtracting 4x to both sides of the equation.

`2x^2 - 4x - 5 = 0`

a = 2

b = -4

c = -5

The quadratic formula is:

`x = (-b +- sqrt (b^2 - 4ac))/(2a)`

Plug in your a, b, and c.

`x = (4 +- sqrt ((-4)^2 - 4(2)(-5)))/(2(2))`

`x = (4+- sqrt (56))/(4)`

**QUESTION:-**

2x squared -5=4x

- Use quadratic formula
- Write the equation in standard form
- Identify a,b,c
- Solve by using quadratic formula
- Identify the solution set

**SOLUTION:-**

`2x^2-5=4x`

`2x^2-4x-5=0`

``

Quadratic Equation:-

`x={-b+-sqrt(b^2-4ac)}/(2a)`

``

Where;

a = 2

b = -4

c = -5

Insert the values of a, b and c in the quadratic Equation:-

`x={-(-4)+-sqrt((-4)^2-4(2)(-5))}/(2*2)`

`x={4+-sqrt(16+40)}/4`

``

`x={4+-sqrt(56)}/4`

Since there is no square root of 56, so it wont be further simplified;

``

` `

Hence the solution set is: `{(4+sqrt(56))/4,(4-sqrt(56))/4}`

``

2x squared - 5 = 4x

Standard form is `ax^(2)+bx+c=0`

Converting the above equation into standard form:

`2x^2 - 4x - 5 = 0`

Since the above is a quadratic equation a quadratic formula can be used to solve it.

a = 2

b = -4

c = -5

Input the values in the formula,

x= [4+sqrt{-4^2-4(2)(-5)}]/2(2) x= [4-sqrt{-4^2-4(2)(-5)}]/2(2)

x= [4+sqrt(16+40)]/4 x= [4-sqrt(16+40)]/4

x= [4+sqrt(56)]/4 x= [4-sqrt(56)]/4

`x= {4+-sqrt(56)}/4`

``

When using the quadratic formula you have to write your equation set equal to zero. If you subtract 4x from both sides you get **2x^2-4x-5=0**. This would be standard form because the standard quadratic is ax^2+bx+c.

a = 2

b = -4

c = -5

The quadratic formula is `(-b+-sqrt(b^(2)-4ac))/(2a)`` `

When you plug in your numbers you get `(4+-sqrt((-4)^(2)-4(2)(-5)))/(2(2))`

When you solve you get x = `(2+sqrt(14))/(2)`and `(2-sqrt(14))/(2)`

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Ok I'm not really sure why it is showing the formulas like that. I will just attach a photo.