Solve (0Cquadxquad)=x^2-7x+12

$$( 0 C \quad x \quad ) = x ^ { 2 } - 7 x + 12$$

Solve for C (complex solution)

$$C\in \mathrm{C}$$

$$x=3\text{ or }x=4$$

Steps for Solving Linear Equation

Anything times zero gives zero.

$$0=x^{2}-7x+12$$

This is false for any C.

$$C\in$$

Solve for C

$$C\in \mathrm{R}$$

$$x=3\text{ or }x=4$$

Steps for Solving Linear Equation

Anything times zero gives zero.

$$0=x^{2}-7x+12$$

This is false for any C.

$$C\in$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using Factoring

Anything times zero gives zero.

$$0=x^{2}-7x+12$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-7x+12=0$$

To solve the equation, factor x^{2}-7x+12 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.

$$a+b=-7 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

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

$$\left(x-4\right)\left(x-3\right)$$

To find equation solutions, solve x-4=0 and x-3=0.

$$x=4 x=3$$

Steps Using Factoring By Grouping

Anything times zero gives zero.

$$0=x^{2}-7x+12$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-7x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

$$\left(x^{2}-4x\right)+\left(-3x+12\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-3\right)$$

To find equation solutions, solve x-4=0 and x-3=0.

$$x=4 x=3$$

Steps Using the Quadratic Formula

Anything times zero gives zero.

$$0=x^{2}-7x+12$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-7x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-7\right)±1}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±1}{2}$$

Now solve the equation x=\frac{7±1}{2} when ± is plus. Add 7 to 1.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=4 x=3$$

Steps for Completing the Square

Anything times zero gives zero.

$$0=x^{2}-7x+12$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-7x+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$