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(3x+5)(x)=240
We move all terms to the left:
(3x+5)(x)-(240)=0
We multiply parentheses
3x^2+5x-240=0
a = 3; b = 5; c = -240;
Δ = b2-4ac
Δ = 52-4·3·(-240)
Δ = 2905
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{-(5)-\sqrt{2905}}{2*3}=\frac{-5-\sqrt{2905}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{2905}}{2*3}=\frac{-5+\sqrt{2905}}{6} $
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