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r^2+8r=7
We move all terms to the left:
r^2+8r-(7)=0
a = 1; b = 8; c = -7;
Δ = b2-4ac
Δ = 82-4·1·(-7)
Δ = 92
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{23}}{2*1}=\frac{-8-2\sqrt{23}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{23}}{2*1}=\frac{-8+2\sqrt{23}}{2} $
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