4x^2+24x=140

Solution for 4x^2+24x=140 equation:

4x^2+24x=140

We move all terms to the left:

4x^2+24x-(140)=0

a = 4; b = 24; c = -140;
Δ = b2-4ac
Δ = 242-4·4·(-140)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16\sqrt{11}}{2*4}=\frac{-24-16\sqrt{11}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16\sqrt{11}}{2*4}=\frac{-24+16\sqrt{11}}{8}
$