2x^2+23x+52=0.

Solution for 2x^2+23x+52=0. equation:

2x^2+23x+52=0.

a = 2; b = 23; c = +52;
Δ = b2-4ac
Δ = 232-4·2·52
Δ = 113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{113}}{2*2}=\frac{-23-\sqrt{113}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{113}}{2*2}=\frac{-23+\sqrt{113}}{4}
$