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4b^2-46b+126=0
a = 4; b = -46; c = +126;
Δ = b2-4ac
Δ = -462-4·4·126
Δ = 100
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{100}=10$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-10}{2*4}=\frac{36}{8} =4+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+10}{2*4}=\frac{56}{8} =7 $
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