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12x^2+160x+16=0
a = 12; b = 160; c = +16;
Δ = b2-4ac
Δ = 1602-4·12·16
Δ = 24832
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24832}=\sqrt{256*97}=\sqrt{256}*\sqrt{97}=16\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-16\sqrt{97}}{2*12}=\frac{-160-16\sqrt{97}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+16\sqrt{97}}{2*12}=\frac{-160+16\sqrt{97}}{24} $
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