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(4x)(3x+30)=180
We move all terms to the left:
(4x)(3x+30)-(180)=0
We multiply parentheses
12x^2+120x-180=0
a = 12; b = 120; c = -180;
Δ = b2-4ac
Δ = 1202-4·12·(-180)
Δ = 23040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{23040}=\sqrt{2304*10}=\sqrt{2304}*\sqrt{10}=48\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-48\sqrt{10}}{2*12}=\frac{-120-48\sqrt{10}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+48\sqrt{10}}{2*12}=\frac{-120+48\sqrt{10}}{24} $
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