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-15x^2+14x-3=0
a = -15; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·(-15)·(-3)
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*-15}=\frac{-18}{-30} =3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*-15}=\frac{-10}{-30} =1/3 $
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