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10x^2+27x-5=0
a = 10; b = 27; c = -5;
Δ = b2-4ac
Δ = 272-4·10·(-5)
Δ = 929
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{929}}{2*10}=\frac{-27-\sqrt{929}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{929}}{2*10}=\frac{-27+\sqrt{929}}{20} $
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