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5q^2-14q-3=0
a = 5; b = -14; c = -3;
Δ = b2-4ac
Δ = -142-4·5·(-3)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*5}=\frac{-2}{10} =-1/5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*5}=\frac{30}{10} =3 $
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