2x^2+52x+18=0

Solution for 2x^2+52x+18=0 equation:

2x^2+52x+18=0

a = 2; b = 52; c = +18;
Δ = b2-4ac
Δ = 522-4·2·18
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-16\sqrt{10}}{2*2}=\frac{-52-16\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+16\sqrt{10}}{2*2}=\frac{-52+16\sqrt{10}}{4}
$