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10a^2+19a-15=0
a = 10; b = 19; c = -15;
Δ = b2-4ac
Δ = 192-4·10·(-15)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-31}{2*10}=\frac{-50}{20} =-2+1/2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+31}{2*10}=\frac{12}{20} =3/5 $
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