(7a2+3a-6)+(2a2-9a-1)=0

Solution for (7a2+3a-6)+(2a2-9a-1)=0 equation:

(7a^2+3a-6)+(2a^2-9a-1)=0

We get rid of parentheses

7a^2+2a^2+3a-9a-6-1=0

We add all the numbers together, and all the variables

9a^2-6a-7=0

a = 9; b = -6; c = -7;
Δ = b2-4ac
Δ = -62-4·9·(-7)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12\sqrt{2}}{2*9}=\frac{6-12\sqrt{2}}{18}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12\sqrt{2}}{2*9}=\frac{6+12\sqrt{2}}{18}
$