x2+(4-x)(3-x)=7

Solution for x2+(4-x)(3-x)=7 equation:

x2+(4-x)(3-x)=7

We move all terms to the left:

x2+(4-x)(3-x)-(7)=0

We add all the numbers together, and all the variables

x2+(-1x+4)(-1x+3)-7=0

We add all the numbers together, and all the variables

x^2+(-1x+4)(-1x+3)-7=0

We multiply parentheses ..

x^2+(+x^2-3x-4x+12)-7=0

We get rid of parentheses

x^2+x^2-3x-4x+12-7=0

We add all the numbers together, and all the variables

2x^2-7x+5=0

a = 2; b = -7; c = +5;
Δ = b2-4ac
Δ = -72-4·2·5
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{-(-7)-3}{2*2}=\frac{4}{4}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*2}=\frac{10}{4}
=2+1/2
$