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c(c+15)=150
We move all terms to the left:
c(c+15)-(150)=0
We multiply parentheses
c^2+15c-150=0
a = 1; b = 15; c = -150;
Δ = b2-4ac
Δ = 152-4·1·(-150)
Δ = 825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{33}}{2*1}=\frac{-15-5\sqrt{33}}{2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{33}}{2*1}=\frac{-15+5\sqrt{33}}{2} $
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