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5x-4x^2+3=0
a = -4; b = 5; c = +3;
Δ = b2-4ac
Δ = 52-4·(-4)·3
Δ = 73
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{-(5)-\sqrt{73}}{2*-4}=\frac{-5-\sqrt{73}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*-4}=\frac{-5+\sqrt{73}}{-8} $
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