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4x^2+36x=160
We move all terms to the left:
4x^2+36x-(160)=0
a = 4; b = 36; c = -160;
Δ = b2-4ac
Δ = 362-4·4·(-160)
Δ = 3856
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3856}=\sqrt{16*241}=\sqrt{16}*\sqrt{241}=4\sqrt{241}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{241}}{2*4}=\frac{-36-4\sqrt{241}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{241}}{2*4}=\frac{-36+4\sqrt{241}}{8} $
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