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2x^2-29x+48=0
a = 2; b = -29; c = +48;
Δ = b2-4ac
Δ = -292-4·2·48
Δ = 457
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{-(-29)-\sqrt{457}}{2*2}=\frac{29-\sqrt{457}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{457}}{2*2}=\frac{29+\sqrt{457}}{4} $
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