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4x^2+35x=49
We move all terms to the left:
4x^2+35x-(49)=0
a = 4; b = 35; c = -49;
Δ = b2-4ac
Δ = 352-4·4·(-49)
Δ = 2009
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2009}=\sqrt{49*41}=\sqrt{49}*\sqrt{41}=7\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-7\sqrt{41}}{2*4}=\frac{-35-7\sqrt{41}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+7\sqrt{41}}{2*4}=\frac{-35+7\sqrt{41}}{8} $
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