(3x^2-5x-62)+(3x^2+7x)=0

Solution for (3x^2-5x-62)+(3x^2+7x)=0 equation:

(3x^2-5x-62)+(3x^2+7x)=0

We get rid of parentheses

3x^2+3x^2-5x+7x-62=0

We add all the numbers together, and all the variables

6x^2+2x-62=0

a = 6; b = 2; c = -62;
Δ = b2-4ac
Δ = 22-4·6·(-62)
Δ = 1492
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:

$\sqrt{\Delta}=\sqrt{1492}=\sqrt{4*373}=\sqrt{4}*\sqrt{373}=2\sqrt{373}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{373}}{2*6}=\frac{-2-2\sqrt{373}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{373}}{2*6}=\frac{-2+2\sqrt{373}}{12}
$