X^2+5x+25=99/4

Solution for X^2+5x+25=99/4 equation:

X^2+5X+25=99/4

We move all terms to the left:

X^2+5X+25-(99/4)=0

We add all the numbers together, and all the variables

X^2+5X+25-(+99/4)=0

We get rid of parentheses

X^2+5X+25-99/4=0

We multiply all the terms by the denominator


X^2*4+5X*4-99+25*4=0

We add all the numbers together, and all the variables

X^2*4+5X*4+1=0

Wy multiply elements

4X^2+20X+1=0

a = 4; b = 20; c = +1;
Δ = b2-4ac
Δ = 202-4·4·1
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{6}}{2*4}=\frac{-20-8\sqrt{6}}{8}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{6}}{2*4}=\frac{-20+8\sqrt{6}}{8}
$