2x+3-5x^2=0

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Solution for 2x+3-5x^2=0 equation: 



2x+3-5x^2=0

a = -5; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-5)·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-5}=\frac{-10}{-10}
=1
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-5}=\frac{6}{-10}
=-3/5
$

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