(3x^2)+10x=216

Solution for (3x^2)+10x=216 equation:

(3x^2)+10x=216

We move all terms to the left:

(3x^2)+10x-(216)=0

a = 3; b = 10; c = -216;
Δ = b2-4ac
Δ = 102-4·3·(-216)
Δ = 2692
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:

$\sqrt{\Delta}=\sqrt{2692}=\sqrt{4*673}=\sqrt{4}*\sqrt{673}=2\sqrt{673}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{673}}{2*3}=\frac{-10-2\sqrt{673}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{673}}{2*3}=\frac{-10+2\sqrt{673}}{6}
$