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x^2+30x=100
We move all terms to the left:
x^2+30x-(100)=0
a = 1; b = 30; c = -100;
Δ = b2-4ac
Δ = 302-4·1·(-100)
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{13}}{2*1}=\frac{-30-10\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{13}}{2*1}=\frac{-30+10\sqrt{13}}{2} $
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