Answer:
[tex]\sf x=2+\sqrt{6},\:x=2-\sqrt{6}[/tex]
Explanation:
[tex]\sf y = x^2 -4x[/tex]
given that y = 2
[tex]\sf x^2 -4x = 2[/tex]
[tex]\sf x^2 -4x -2 = 0[/tex]
using quadratic formula:
[tex]\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ \ when \ ax^2 + bx + c = 0[/tex]
[tex]\rightarrow \sf x_{1,\:2}=\dfrac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-2\right)}}{2\cdot \:1}[/tex]
[tex]\rightarrow \sf x_1=\dfrac{-\left(-4\right)+2\sqrt{6}}{2\cdot \:1},\:x_2=\dfrac{-\left(-4\right)-2\sqrt{6}}{2\cdot \:1}[/tex]
[tex]\rightarrow \sf x=2+\sqrt{6},\:x=2-\sqrt{6}[/tex]
in decimals:
[tex]\rightarrow \sf x=4.45 , \ -0.45[/tex]
