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(x)(4x+5)=440
We move all terms to the left:
(x)(4x+5)-(440)=0
We multiply parentheses
4x^2+5x-440=0
a = 4; b = 5; c = -440;
Δ = b2-4ac
Δ = 52-4·4·(-440)
Δ = 7065
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7065}=\sqrt{9*785}=\sqrt{9}*\sqrt{785}=3\sqrt{785}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{785}}{2*4}=\frac{-5-3\sqrt{785}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{785}}{2*4}=\frac{-5+3\sqrt{785}}{8} $
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