1+3x-2x^2=0

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Solution for 1+3x-2x^2=0 equation: 



1+3x-2x^2=0

a = -2; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·(-2)·1
Δ = 17
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{17}}{2*-2}=\frac{-3-\sqrt{17}}{-4}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{17}}{2*-2}=\frac{-3+\sqrt{17}}{-4}
$

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