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15x^2+x-6=0
a = 15; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·15·(-6)
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*15}=\frac{-20}{30} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*15}=\frac{18}{30} =3/5 $
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