# How do I solve the equation 5x^2=8x+2 using the quadratic formula?

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The roots of a quadratic equation ax^2 + bx + c = 0 are equal to

x1 = [-b + sqrt (b^2 - 4ac)]/2a and x2 = [-b - sqrt (b^2 - 4ac)]/2a

Here the equation we have is 5x^2 = 8x + 2

=> 5x^2 - 8x - 2 = 0

a = 5, b = -8 and c = -2

x1 = [8+ sqrt (64 + 40)]/10

=> 8/10 + sqrt 104 / 10

=> 4/5 + sqrt 26 / 5

x2 = 4/5 - sqrt 26 / 5

**The solutions of the equation are x = 4/5 + sqrt 26 / 5 and x = 4/5 - sqrt 26 / 5**

`5x^2=8x+2 `

move the numbers on the same side

`5x^2-8x-2 `

a=5 b=-8 c=-2

use the quadratic formula

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

plug in the numbers

`(8+-sqrt(-8^2-4(5)(-2)))/ (2(5)) `

`(8+-sqrt(64+40))/ (10) `

`(8+-sqrt(64+40))/ (10) `

`(8+-sqrt(104))/ (10) `

`x=(8-sqrt(104))/ (10) `

`x=(8+sqrt(104))/ (10)`

To solve an equation means to determine it's roots. Since it is a quadratic equation, it will have 2 roots.

The quadratic formula is used to determine the roots:

x1 = [-b+sqrt(delta)]/2a and x2 = [-b-sqrt(delta)]/2a

discriminant = delta = b^2 - 4ac

We'll identify the coefficient of the quadratic: a,b,c.

a = 5, b = -8 and c = -2

delta = (-8)^2 - 4*5*(-2)

delta = 64 + 40

delta = 104

We'll take square root both sides:

sqrt delta = sqrt 104

sqrt delta = 2sqrt 26

x1 = (8 + 2sqrt 26)/2*5

x1 = (4 + sqrt26)/5

x2 = (4 - sqrt26)/5

**The roots of the quadratic are: x1 = (4 + sqrt26)/5 and x2 = (4 - sqrt26)/5.**