3x^2+20x=62

Solution for 3x^2+20x=62 equation:

3x^2+20x=62

We move all terms to the left:

3x^2+20x-(62)=0

a = 3; b = 20; c = -62;
Δ = b2-4ac
Δ = 202-4·3·(-62)
Δ = 1144
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:

$\sqrt{\Delta}=\sqrt{1144}=\sqrt{4*286}=\sqrt{4}*\sqrt{286}=2\sqrt{286}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{286}}{2*3}=\frac{-20-2\sqrt{286}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{286}}{2*3}=\frac{-20+2\sqrt{286}}{6}
$