10x^2+20x-89=9

Solution for 10x^2+20x-89=9 equation:

10x^2+20x-89=9

We move all terms to the left:

10x^2+20x-89-(9)=0

We add all the numbers together, and all the variables

10x^2+20x-98=0

a = 10; b = 20; c = -98;
Δ = b2-4ac
Δ = 202-4·10·(-98)
Δ = 4320
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:

$\sqrt{\Delta}=\sqrt{4320}=\sqrt{144*30}=\sqrt{144}*\sqrt{30}=12\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{30}}{2*10}=\frac{-20-12\sqrt{30}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{30}}{2*10}=\frac{-20+12\sqrt{30}}{20}
$