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2b^2-10b+12=0
a = 2; b = -10; c = +12;
Δ = b2-4ac
Δ = -102-4·2·12
Δ = 4
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{4}=2$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*2}=\frac{8}{4} =2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*2}=\frac{12}{4} =3 $
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