X^2+3x+(9/4)=(41/4)

Solution for X^2+3x+(9/4)=(41/4) equation:

X^2+3X+(9/4)=(41/4)

We move all terms to the left:

X^2+3X+(9/4)-((41/4))=0

We add all the numbers together, and all the variables

X^2+3X+(+9/4)-((+41/4))=0

We get rid of parentheses

X^2+3X+9/4-((+41/4))=0

We calculate fractions


X^2+3X=0

a = 1; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·1·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*1}=\frac{-6}{2}
=-3
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*1}=\frac{0}{2}
=0
$