3x^2+6x=35

Solution for 3x^2+6x=35 equation:

3x^2+6x=35

We move all terms to the left:

3x^2+6x-(35)=0

a = 3; b = 6; c = -35;
Δ = b2-4ac
Δ = 62-4·3·(-35)
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{114}}{2*3}=\frac{-6-2\sqrt{114}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{114}}{2*3}=\frac{-6+2\sqrt{114}}{6}
$