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5r^2+5r-12=0
a = 5; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·5·(-12)
Δ = 265
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{-(5)-\sqrt{265}}{2*5}=\frac{-5-\sqrt{265}}{10} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{265}}{2*5}=\frac{-5+\sqrt{265}}{10} $
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