`2x^2 + 3x = 20` Solve with the quadratic formula.

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lemjay | High School Teacher | (Level 3) Senior Educator

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A quadratic equation in a form ax^2+bx+c=0 can be solved using the quadratic formula which is:

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

For the problem above, to be able to apply the quadratic formula, set one side equation equal to zero.

`2x^2+3x=20`

`2x^2+3x-20=20-20`

`2x^2+3x-20=0`

Now that one side of the equation is zero, plug-in the values of a, b and c to the formula. The values are a=2, b=3 and c=-20.

`x=(-3+-sqrt(3^2-4(2)(-20)))/(2*2)`

`x=(-3+-sqrt(9+160))/4`

`x=(-3+-sqrt169)/4`

`x=(-3+-13)/4`

Since the numerator has `+- ` between 3 and 13, the x breaks into two.

`x=(-3+13)/4=10/4=5/2`     and     `x=(-3-13)/4=(-16)/4=-4`

Hence, the solution to `2x^2+3x=20` is `x={-4,5/2}` .

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oldnick | (Level 1) Valedictorian

Posted on

`2x^2+3x=20`

`2x^2+3x-20=0 `

`Delta=9-4(2)(-20)=9+160=169`

Since  `Delta >0`  equation has two different solutions:

`x=(-3+-sqrt(169))/4` `=(-3+-13)/4`      `x_1=5/2`   `x_2=-4`   

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