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4x^2-49x+16=0
a = 4; b = -49; c = +16;
Δ = b2-4ac
Δ = -492-4·4·16
Δ = 2145
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-\sqrt{2145}}{2*4}=\frac{49-\sqrt{2145}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2145}}{2*4}=\frac{49+\sqrt{2145}}{8} $
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