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60x+4x^2=396
We move all terms to the left:
60x+4x^2-(396)=0
a = 4; b = 60; c = -396;
Δ = b2-4ac
Δ = 602-4·4·(-396)
Δ = 9936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9936}=\sqrt{144*69}=\sqrt{144}*\sqrt{69}=12\sqrt{69}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-12\sqrt{69}}{2*4}=\frac{-60-12\sqrt{69}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+12\sqrt{69}}{2*4}=\frac{-60+12\sqrt{69}}{8} $
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