2x^2+12x-53=0

Solution for 2x^2+12x-53=0 equation:

2x^2+12x-53=0

a = 2; b = 12; c = -53;
Δ = b2-4ac
Δ = 122-4·2·(-53)
Δ = 568
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:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{142}}{2*2}=\frac{-12-2\sqrt{142}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{142}}{2*2}=\frac{-12+2\sqrt{142}}{4}
$