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r^2+7r-12=0
a = 1; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·1·(-12)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{97}}{2*1}=\frac{-7-\sqrt{97}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{97}}{2*1}=\frac{-7+\sqrt{97}}{2} $
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