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(x)(2x+3)=440
We move all terms to the left:
(x)(2x+3)-(440)=0
We multiply parentheses
2x^2+3x-440=0
a = 2; b = 3; c = -440;
Δ = b2-4ac
Δ = 32-4·2·(-440)
Δ = 3529
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{3529}}{2*2}=\frac{-3-\sqrt{3529}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{3529}}{2*2}=\frac{-3+\sqrt{3529}}{4} $
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