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3b^2-3b-10=0
a = 3; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·3·(-10)
Δ = 129
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{129}}{2*3}=\frac{3-\sqrt{129}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*3}=\frac{3+\sqrt{129}}{6} $
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