4x^2+3-7x=0

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Solution for 4x^2+3-7x=0 equation: 



4x^2+3-7x=0

a = 4; b = -7; c = +3;
Δ = b2-4ac
Δ = -72-4·4·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*4}=\frac{6}{8}
=3/4
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*4}=\frac{8}{8}
=1
$

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