Solve 10-3x+x^2-7-4x=

$$10 - 3 x + x ^ { 2 } - 7 - 4 x =$$

Evaluate

$$x^{2}-7x+3$$

Solution Steps

Subtract 7 from 10 to get 3.

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

Combine -3x and -4x to get -7x.

$$3-7x+x^{2}$$

Factor

$$\left(x-\frac{7-\sqrt{37}}{2}\right)\left(x-\frac{\sqrt{37}+7}{2}\right)$$

Steps Using the Quadratic Formula

Subtract 7 from 10 to get 3.

$$factor(3-3x+x^{2}-4x)$$

Combine -3x and -4x to get -7x.

$$factor(3-7x+x^{2})$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-7x+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3}}{2}$$

Multiply -4 times 3.

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

Add 49 to -12.

$$x=\frac{-\left(-7\right)±\sqrt{37}}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{37}}{2}$$

Now solve the equation x=\frac{7±\sqrt{37}}{2} when ± is plus. Add 7 to \sqrt{37}.

$$x=\frac{\sqrt{37}+7}{2}$$

Now solve the equation x=\frac{7±\sqrt{37}}{2} when ± is minus. Subtract \sqrt{37} from 7.

$$x=\frac{7-\sqrt{37}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7+\sqrt{37}}{2} for x_{1} and \frac{7-\sqrt{37}}{2} for x_{2}.

$$x^{2}-7x+3=\left(x-\frac{\sqrt{37}+7}{2}\right)\left(x-\frac{7-\sqrt{37}}{2}\right)$$