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4x^2+35x-144=0
a = 4; b = 35; c = -144;
Δ = b2-4ac
Δ = 352-4·4·(-144)
Δ = 3529
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{3529}}{2*4}=\frac{-35-\sqrt{3529}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{3529}}{2*4}=\frac{-35+\sqrt{3529}}{8} $
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