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b^2-24b+140=0
a = 1; b = -24; c = +140;
Δ = b2-4ac
Δ = -242-4·1·140
Δ = 16
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{16}=4$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4}{2*1}=\frac{20}{2} =10 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4}{2*1}=\frac{28}{2} =14 $
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