2x^2+142=-36x

Solution for 2x^2+142=-36x equation:

2x^2+142=-36x

We move all terms to the left:

2x^2+142-(-36x)=0

We get rid of parentheses

2x^2+36x+142=0

a = 2; b = 36; c = +142;
Δ = b2-4ac
Δ = 362-4·2·142
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{10}}{2*2}=\frac{-36-4\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{10}}{2*2}=\frac{-36+4\sqrt{10}}{4}
$