2x(3x+20)=180

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Solution for 2x(3x+20)=180 equation:



2x(3x+20)=180
We move all terms to the left:
2x(3x+20)-(180)=0
We multiply parentheses
6x^2+40x-180=0
a = 6; b = 40; c = -180;
Δ = b2-4ac
Δ = 402-4·6·(-180)
Δ = 5920
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:
$\sqrt{\Delta}=\sqrt{5920}=\sqrt{16*370}=\sqrt{16}*\sqrt{370}=4\sqrt{370}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{370}}{2*6}=\frac{-40-4\sqrt{370}}{12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{370}}{2*6}=\frac{-40+4\sqrt{370}}{12} $

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