4x^2+10x=96

Solution for 4x^2+10x=96 equation:

4x^2+10x=96

We move all terms to the left:

4x^2+10x-(96)=0

a = 4; b = 10; c = -96;
Δ = b2-4ac
Δ = 102-4·4·(-96)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{409}}{2*4}=\frac{-10-2\sqrt{409}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{409}}{2*4}=\frac{-10+2\sqrt{409}}{8}
$