4x^2+18+112=180

Solution for 4x^2+18+112=180 equation:

4x^2+18+112=180

We move all terms to the left:

4x^2+18+112-(180)=0

We add all the numbers together, and all the variables

4x^2-50=0

a = 4; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·4·(-50)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*4}=\frac{0-20\sqrt{2}}{8}
=-\frac{20\sqrt{2}}{8}
=-\frac{5\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*4}=\frac{0+20\sqrt{2}}{8}
=\frac{20\sqrt{2}}{8}
=\frac{5\sqrt{2}}{2}
$