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x^2+64x+120=0
a = 1; b = 64; c = +120;
Δ = b2-4ac
Δ = 642-4·1·120
Δ = 3616
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3616}=\sqrt{16*226}=\sqrt{16}*\sqrt{226}=4\sqrt{226}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-4\sqrt{226}}{2*1}=\frac{-64-4\sqrt{226}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+4\sqrt{226}}{2*1}=\frac{-64+4\sqrt{226}}{2} $
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