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2x^2+5x=440
We move all terms to the left:
2x^2+5x-(440)=0
a = 2; b = 5; c = -440;
Δ = b2-4ac
Δ = 52-4·2·(-440)
Δ = 3545
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{3545}}{2*2}=\frac{-5-\sqrt{3545}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{3545}}{2*2}=\frac{-5+\sqrt{3545}}{4} $
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