4x^2+64x+167=2x^2+42x+107

Solution for 4x^2+64x+167=2x^2+42x+107 equation:

4x^2+64x+167=2x^2+42x+107

We move all terms to the left:

4x^2+64x+167-(2x^2+42x+107)=0

We get rid of parentheses

4x^2-2x^2+64x-42x-107+167=0

We add all the numbers together, and all the variables

2x^2+22x+60=0

a = 2; b = 22; c = +60;
Δ = b2-4ac
Δ = 222-4·2·60
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*2}=\frac{-20}{4}
=-5
$