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5x^2+11x-10=0
a = 5; b = 11; c = -10;
Δ = b2-4ac
Δ = 112-4·5·(-10)
Δ = 321
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{-(11)-\sqrt{321}}{2*5}=\frac{-11-\sqrt{321}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{321}}{2*5}=\frac{-11+\sqrt{321}}{10} $
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