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(4x+3)x=240
We move all terms to the left:
(4x+3)x-(240)=0
We multiply parentheses
4x^2+3x-240=0
a = 4; b = 3; c = -240;
Δ = b2-4ac
Δ = 32-4·4·(-240)
Δ = 3849
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{3849}}{2*4}=\frac{-3-\sqrt{3849}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{3849}}{2*4}=\frac{-3+\sqrt{3849}}{8} $
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