2x^2+20x+51=9

Solution for 2x^2+20x+51=9 equation:

2x^2+20x+51=9

We move all terms to the left:

2x^2+20x+51-(9)=0

We add all the numbers together, and all the variables

2x^2+20x+42=0

a = 2; b = 20; c = +42;
Δ = b2-4ac
Δ = 202-4·2·42
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8}{2*2}=\frac{-28}{4}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8}{2*2}=\frac{-12}{4}
=-3
$