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7r^2+32r+16=0
a = 7; b = 32; c = +16;
Δ = b2-4ac
Δ = 322-4·7·16
Δ = 576
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-24}{2*7}=\frac{-56}{14} =-4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+24}{2*7}=\frac{-8}{14} =-4/7 $
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