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x^2-8x=42
We move all terms to the left:
x^2-8x-(42)=0
a = 1; b = -8; c = -42;
Δ = b2-4ac
Δ = -82-4·1·(-42)
Δ = 232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{58}}{2*1}=\frac{8-2\sqrt{58}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{58}}{2*1}=\frac{8+2\sqrt{58}}{2} $
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