(8X-3)/4=x^2

Solution for (8X-3)/4=x^2 equation:

(8X-3)/4=X^2

We move all terms to the left:

(8X-3)/4-(X^2)=0

determiningTheFunctionDomain
-X^2+(8X-3)/4=0

We add all the numbers together, and all the variables

-1X^2+(8X-3)/4=0

We multiply all the terms by the denominator


-1X^2*4+(8X-3)=0

Wy multiply elements

-4X^2+(8X-3)=0

We get rid of parentheses

-4X^2+8X-3=0

a = -4; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·(-4)·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-4}=\frac{-4}{-8}
=1/2
$