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4x^2+29x-49=0
a = 4; b = 29; c = -49;
Δ = b2-4ac
Δ = 292-4·4·(-49)
Δ = 1625
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-5\sqrt{65}}{2*4}=\frac{-29-5\sqrt{65}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+5\sqrt{65}}{2*4}=\frac{-29+5\sqrt{65}}{8} $
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