4x^2+12x=35

Solution for 4x^2+12x=35 equation:

4x^2+12x=35

We move all terms to the left:

4x^2+12x-(35)=0

a = 4; b = 12; c = -35;
Δ = b2-4ac
Δ = 122-4·4·(-35)
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{11}}{2*4}=\frac{-12-8\sqrt{11}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{11}}{2*4}=\frac{-12+8\sqrt{11}}{8}
$