x^2+5x+(25/4)=3/4

Solution for x^2+5x+(25/4)=3/4 equation:

x^2+5x+(25/4)=3/4

We move all terms to the left:

x^2+5x+(25/4)-(3/4)=0

We add all the numbers together, and all the variables

x^2+5x+(+25/4)-(+3/4)=0

We get rid of parentheses

x^2+5x+25/4-3/4=0

We multiply all the terms by the denominator


x^2*4+5x*4+25-3=0

We add all the numbers together, and all the variables

x^2*4+5x*4+22=0

Wy multiply elements

4x^2+20x+22=0

a = 4; b = 20; c = +22;
Δ = b2-4ac
Δ = 202-4·4·22
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{3}}{2*4}=\frac{-20-4\sqrt{3}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{3}}{2*4}=\frac{-20+4\sqrt{3}}{8}
$