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4x^2+3x-1.69=0
a = 4; b = 3; c = -1.69;
Δ = b2-4ac
Δ = 32-4·4·(-1.69)
Δ = 36.04
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{-(3)-\sqrt{36.04}}{2*4}=\frac{-3-\sqrt{36.04}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{36.04}}{2*4}=\frac{-3+\sqrt{36.04}}{8} $
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