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6b-10b^2=0
a = -10; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-10)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-10}=\frac{-12}{-20} =3/5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-10}=\frac{0}{-20} =0 $
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