x^2+(3x/2)=7

Solution for x^2+(3x/2)=7 equation:

x^2+(3x/2)=7

We move all terms to the left:

x^2+(3x/2)-(7)=0

We add all the numbers together, and all the variables

x^2+(+3x/2)-7=0

We get rid of parentheses

x^2+3x/2-7=0

We multiply all the terms by the denominator


x^2*2+3x-7*2=0

We add all the numbers together, and all the variables

x^2*2+3x-14=0

Wy multiply elements

2x^2+3x-14=0

a = 2; b = 3; c = -14;
Δ = b2-4ac
Δ = 32-4·2·(-14)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*2}=\frac{-14}{4}
=-3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*2}=\frac{8}{4}
=2
$