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4x^2+98+45x=0
a = 4; b = 45; c = +98;
Δ = b2-4ac
Δ = 452-4·4·98
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{-(45)-\sqrt{457}}{2*4}=\frac{-45-\sqrt{457}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{457}}{2*4}=\frac{-45+\sqrt{457}}{8} $
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