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10x^2+53x+15=0
a = 10; b = 53; c = +15;
Δ = b2-4ac
Δ = 532-4·10·15
Δ = 2209
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{2209}=47$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-47}{2*10}=\frac{-100}{20} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+47}{2*10}=\frac{-6}{20} =-3/10 $
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