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4x^2+24x+35=0
a = 4; b = 24; c = +35;
Δ = b2-4ac
Δ = 242-4·4·35
Δ = 16
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}\sqrt{\Delta}=\sqrt{16}=4x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4}{2*4}=\frac{-28}{8} =-3+1/2x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4}{2*4}=\frac{-20}{8} =-2+1/2
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