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10x+x^2=68
We move all terms to the left:
10x+x^2-(68)=0
a = 1; b = 10; c = -68;
Δ = b2-4ac
Δ = 102-4·1·(-68)
Δ = 372
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{93}}{2*1}=\frac{-10-2\sqrt{93}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{93}}{2*1}=\frac{-10+2\sqrt{93}}{2} $
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