Answer :
[tex]x^{2}+12=7*x[/tex], where x is unknown numer
Now we are moving [tex]7*x[/tex] on left side to obtain quadratic equation:
[tex]x^{2}+12-7*x=0[/tex]
Our goal is to find two roots of this equation.
First we are finding delta:
[tex]\Delta =b^{2}-4*a*c[/tex]
[tex]\Delta =(-7)^{2}-4*1*12=1[/tex]
First root:
[tex]x_{1}=\frac{-b+\sqrt{\Delta}}{2*a}[/tex]
[tex]x_{1}=\frac{7+\sqrt{1 }}{2}=4[/tex]
Second root:
[tex]x_{2}=\frac{-b-\sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2}=\frac{7-\sqrt{1 }}{2}=3[/tex]
Now we are moving [tex]7*x[/tex] on left side to obtain quadratic equation:
[tex]x^{2}+12-7*x=0[/tex]
Our goal is to find two roots of this equation.
First we are finding delta:
[tex]\Delta =b^{2}-4*a*c[/tex]
[tex]\Delta =(-7)^{2}-4*1*12=1[/tex]
First root:
[tex]x_{1}=\frac{-b+\sqrt{\Delta}}{2*a}[/tex]
[tex]x_{1}=\frac{7+\sqrt{1 }}{2}=4[/tex]
Second root:
[tex]x_{2}=\frac{-b-\sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2}=\frac{7-\sqrt{1 }}{2}=3[/tex]
[tex]x^2+12=7x\\
x^2-7x+12=0\\
x^2-3x-4x+12=0\\
x(x-3)-4(x-3)=0\\
(x-4)(x-3)=0\\
x=4 \vee x=3[/tex]