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5x^2+27x+10=0
a = 5; b = 27; c = +10;
Δ = b2-4ac
Δ = 272-4·5·10
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-23}{2*5}=\frac{-50}{10} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+23}{2*5}=\frac{-4}{10} =-2/5 $
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