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4x^2+14=35x
We move all terms to the left:
4x^2+14-(35x)=0
a = 4; b = -35; c = +14;
Δ = b2-4ac
Δ = -352-4·4·14
Δ = 1001
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{-(-35)-\sqrt{1001}}{2*4}=\frac{35-\sqrt{1001}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1001}}{2*4}=\frac{35+\sqrt{1001}}{8} $
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