How do you solve $2x^2 + 12x= 10$ using the quadratic formula?

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Answer 1 · 2017-05-09T16:14:24

$x=-3+-sqrt14$

first rearrange to the form $" "ax^2+bx+c=0$

we have

$2x^2+12x-10=0$

divide out any common factors to make it simpler to use.

$(2x^2+12x-10=0)-:2$

$=>x^2+6x-5=0$

the formula is

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

$a=1, b=6, c=-5$

$x=(-6+-sqrt(6^2-4xx1xx(-5)))/(2xx1)$

$x=(-6+-sqrt(36+20))/2$

$x=(-6+-sqrt56)/2$

$x=-6/2+-sqrt56/2$

$x=-6/2+-2sqrt14/2$

$:.x=-3+-sqrt14$

How do you solve 2x^2 + 12x= 10 2x^2 + 12x= 10 using the quadratic formula?