Solve 7-5x=2x^2

$$7 - 5 x = 2 x ^ { 2 }$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$7-5x-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}-5x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-2\times 7=-14$$

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. List all such integer pairs that give product -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=2 b=-7$$

Rewrite -2x^{2}-5x+7 as \left(-2x^{2}+2x\right)+\left(-7x+7\right).

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

Factor out 2x in the first and 7 in the second group.

$$2x\left(-x+1\right)+7\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(2x+7\right)$$

To find equation solutions, solve -x+1=0 and 2x+7=0.

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

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$7-5x-2x^{2}=0$$

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.

$$-2x^{2}-5x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 7}}{2\left(-2\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 7}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 7}}{2\left(-2\right)}$$

Multiply 8 times 7.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2\left(-2\right)}$$

Add 25 to 56.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2\left(-2\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-5\right)±9}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±9}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±9}{-4}$$

Now solve the equation x=\frac{5±9}{-4} when ± is plus. Add 5 to 9.

$$x=\frac{14}{-4}$$

Reduce the fraction \frac{14}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{5±9}{-4} when ± is minus. Subtract 9 from 5.

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

Divide -4 by -4.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$7-5x-2x^{2}=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$-5x-2x^{2}=-7$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}-5x=-7$$

Divide both sides by -2.

$$\frac{-2x^{2}-5x}{-2}=-\frac{7}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{5}{-2}\right)x=-\frac{7}{-2}$$

Divide -5 by -2.

$$x^{2}+\frac{5}{2}x=-\frac{7}{-2}$$

Divide -7 by -2.

$$x^{2}+\frac{5}{2}x=\frac{7}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{7}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{7}{2}+\frac{25}{16}$$

Add \frac{7}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{9}{4} x+\frac{5}{4}=-\frac{9}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

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