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7r^2-40r-12=0
a = 7; b = -40; c = -12;
Δ = b2-4ac
Δ = -402-4·7·(-12)
Δ = 1936
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{1936}=44$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-44}{2*7}=\frac{-4}{14} =-2/7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+44}{2*7}=\frac{84}{14} =6 $
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