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2x^2+25x+4=0
a = 2; b = 25; c = +4;
Δ = b2-4ac
Δ = 252-4·2·4
Δ = 593
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{-(25)-\sqrt{593}}{2*2}=\frac{-25-\sqrt{593}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{593}}{2*2}=\frac{-25+\sqrt{593}}{4} $
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