Solve x^2-4x+2=-3x+4

$$x ^ { 2 } - 4 x + 2 = - 3 x + 4$$

Solve for x

$$x=-1$$

$$x=2$$

Steps Using Factoring

Add 3x to both sides.

$$x^{2}-4x+2+3x=4$$

Combine -4x and 3x to get -x.

$$x^{2}-x+2=4$$

Subtract 4 from both sides.

$$x^{2}-x+2-4=0$$

Subtract 4 from 2 to get -2.

$$x^{2}-x-2=0$$

To solve the equation, factor x^{2}-x-2 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Steps Using Factoring By Grouping

Add 3x to both sides.

$$x^{2}-4x+2+3x=4$$

Combine -4x and 3x to get -x.

$$x^{2}-x+2=4$$

Subtract 4 from both sides.

$$x^{2}-x+2-4=0$$

Subtract 4 from 2 to get -2.

$$x^{2}-x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite x^{2}-x-2 as \left(x^{2}-2x\right)+\left(x-2\right).

$$\left(x^{2}-2x\right)+\left(x-2\right)$$

Factor out x in x^{2}-2x.

$$x\left(x-2\right)+x-2$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Steps Using the Quadratic Formula

Add 3x to both sides.

$$x^{2}-4x+2+3x=4$$

Combine -4x and 3x to get -x.

$$x^{2}-x+2=4$$

Subtract 4 from both sides.

$$x^{2}-x+2-4=0$$

Subtract 4 from 2 to get -2.

$$x^{2}-x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2}$$

Add 1 to 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-1\right)±3}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±3}{2}$$

Now solve the equation x=\frac{1±3}{2} when ± is plus. Add 1 to 3.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.

$$x=-\frac{2}{2}$$

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=2 x=-1$$

Steps for Completing the Square

Add 3x to both sides.

$$x^{2}-4x+2+3x=4$$

Combine -4x and 3x to get -x.

$$x^{2}-x+2=4$$

Subtract 2 from both sides.

$$x^{2}-x=4-2$$

Subtract 2 from 4 to get 2.

$$x^{2}-x=2$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=2+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$