2x^2+10x-48=720

Solution for 2x^2+10x-48=720 equation:

2x^2+10x-48=720

We move all terms to the left:

2x^2+10x-48-(720)=0

We add all the numbers together, and all the variables

2x^2+10x-768=0

a = 2; b = 10; c = -768;
Δ = b2-4ac
Δ = 102-4·2·(-768)
Δ = 6244
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:

$\sqrt{\Delta}=\sqrt{6244}=\sqrt{4*1561}=\sqrt{4}*\sqrt{1561}=2\sqrt{1561}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1561}}{2*2}=\frac{-10-2\sqrt{1561}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1561}}{2*2}=\frac{-10+2\sqrt{1561}}{4}
$