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10x^2+27x-9=0
a = 10; b = 27; c = -9;
Δ = b2-4ac
Δ = 272-4·10·(-9)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-33}{2*10}=\frac{-60}{20} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+33}{2*10}=\frac{6}{20} =3/10 $
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