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8r^2+16r-24=0
a = 8; b = 16; c = -24;
Δ = b2-4ac
Δ = 162-4·8·(-24)
Δ = 1024
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{1024}=32$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*8}=\frac{-48}{16} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*8}=\frac{16}{16} =1 $
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