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4x^2+9x=440
We move all terms to the left:
4x^2+9x-(440)=0
a = 4; b = 9; c = -440;
Δ = b2-4ac
Δ = 92-4·4·(-440)
Δ = 7121
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{-(9)-\sqrt{7121}}{2*4}=\frac{-9-\sqrt{7121}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{7121}}{2*4}=\frac{-9+\sqrt{7121}}{8} $
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