2a^2+4a+5=7a^2-9a-8

Solution for 2a^2+4a+5=7a^2-9a-8 equation:

2a^2+4a+5=7a^2-9a-8

We move all terms to the left:

2a^2+4a+5-(7a^2-9a-8)=0

We get rid of parentheses

2a^2-7a^2+4a+9a+8+5=0

We add all the numbers together, and all the variables

-5a^2+13a+13=0

a = -5; b = 13; c = +13;
Δ = b2-4ac
Δ = 132-4·(-5)·13
Δ = 429
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{429}}{2*-5}=\frac{-13-\sqrt{429}}{-10}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{429}}{2*-5}=\frac{-13+\sqrt{429}}{-10}
$