4x^2+24x=35

Solution for 4x^2+24x=35 equation:

4x^2+24x=35

We move all terms to the left:

4x^2+24x-(35)=0

a = 4; b = 24; c = -35;
Δ = b2-4ac
Δ = 242-4·4·(-35)
Δ = 1136
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:

$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{71}}{2*4}=\frac{-24-4\sqrt{71}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{71}}{2*4}=\frac{-24+4\sqrt{71}}{8}
$