2x^2+21x+54=180

Solution for 2x^2+21x+54=180 equation:

2x^2+21x+54=180

We move all terms to the left:

2x^2+21x+54-(180)=0

We add all the numbers together, and all the variables

2x^2+21x-126=0

a = 2; b = 21; c = -126;
Δ = b2-4ac
Δ = 212-4·2·(-126)
Δ = 1449
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:

$\sqrt{\Delta}=\sqrt{1449}=\sqrt{9*161}=\sqrt{9}*\sqrt{161}=3\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{161}}{2*2}=\frac{-21-3\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{161}}{2*2}=\frac{-21+3\sqrt{161}}{4}
$