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4x^2-40(x)+51=0
a = 4; b = -40; c = +51;
Δ = b2-4ac
Δ = -402-4·4·51
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-28}{2*4}=\frac{12}{8} =1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+28}{2*4}=\frac{68}{8} =8+1/2 $
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