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(3x+40)(4x)=180
We move all terms to the left:
(3x+40)(4x)-(180)=0
We multiply parentheses
12x^2+160x-180=0
a = 12; b = 160; c = -180;
Δ = b2-4ac
Δ = 1602-4·12·(-180)
Δ = 34240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{34240}=\sqrt{64*535}=\sqrt{64}*\sqrt{535}=8\sqrt{535}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{535}}{2*12}=\frac{-160-8\sqrt{535}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{535}}{2*12}=\frac{-160+8\sqrt{535}}{24} $
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