(7x^2-7x+21)+(8x^2+9x-42)=0

Solution for (7x^2-7x+21)+(8x^2+9x-42)=0 equation:

(7x^2-7x+21)+(8x^2+9x-42)=0

We get rid of parentheses

7x^2+8x^2-7x+9x+21-42=0

We add all the numbers together, and all the variables

15x^2+2x-21=0

a = 15; b = 2; c = -21;
Δ = b2-4ac
Δ = 22-4·15·(-21)
Δ = 1264
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:

$\sqrt{\Delta}=\sqrt{1264}=\sqrt{16*79}=\sqrt{16}*\sqrt{79}=4\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{79}}{2*15}=\frac{-2-4\sqrt{79}}{30}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{79}}{2*15}=\frac{-2+4\sqrt{79}}{30}
$