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4x^2+26x-55=0
a = 4; b = 26; c = -55;
Δ = b2-4ac
Δ = 262-4·4·(-55)
Δ = 1556
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1556}=\sqrt{4*389}=\sqrt{4}*\sqrt{389}=2\sqrt{389}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{389}}{2*4}=\frac{-26-2\sqrt{389}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{389}}{2*4}=\frac{-26+2\sqrt{389}}{8} $
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