Solve 7x^2=45x+28

$$7 x ^ { 2 } = 45 x + 28$$

Solve for x

$$x=-\frac{4}{7}\approx -0.571428571$$

$$x=7$$

Steps Using Factoring By Grouping

Subtract 45x from both sides.

$$7x^{2}-45x=28$$

Subtract 28 from both sides.

$$7x^{2}-45x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=-45 ab=7\left(-28\right)=-196$$

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 -196.

$$1,-196 2,-98 4,-49 7,-28 14,-14$$

Calculate the sum for each pair.

$$1-196=-195 2-98=-96 4-49=-45 7-28=-21 14-14=0$$

The solution is the pair that gives sum -45.

$$a=-49 b=4$$

Rewrite 7x^{2}-45x-28 as \left(7x^{2}-49x\right)+\left(4x-28\right).

$$\left(7x^{2}-49x\right)+\left(4x-28\right)$$

Factor out 7x in the first and 4 in the second group.

$$7x\left(x-7\right)+4\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(7x+4\right)$$

To find equation solutions, solve x-7=0 and 7x+4=0.

$$x=7 x=-\frac{4}{7}$$

Steps Using the Quadratic Formula

Subtract 45x from both sides.

$$7x^{2}-45x=28$$

Subtract 28 from both sides.

$$7x^{2}-45x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -45 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-45\right)±\sqrt{\left(-45\right)^{2}-4\times 7\left(-28\right)}}{2\times 7}$$

Square -45.

$$x=\frac{-\left(-45\right)±\sqrt{2025-4\times 7\left(-28\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-45\right)±\sqrt{2025-28\left(-28\right)}}{2\times 7}$$

Multiply -28 times -28.

$$x=\frac{-\left(-45\right)±\sqrt{2025+784}}{2\times 7}$$

Add 2025 to 784.

$$x=\frac{-\left(-45\right)±\sqrt{2809}}{2\times 7}$$

Take the square root of 2809.

$$x=\frac{-\left(-45\right)±53}{2\times 7}$$

The opposite of -45 is 45.

$$x=\frac{45±53}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{45±53}{14}$$

Now solve the equation x=\frac{45±53}{14} when ± is plus. Add 45 to 53.

$$x=\frac{98}{14}$$

Divide 98 by 14.

$$x=7$$

Now solve the equation x=\frac{45±53}{14} when ± is minus. Subtract 53 from 45.

$$x=-\frac{8}{14}$$

Reduce the fraction \frac{-8}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{7}$$

The equation is now solved.

$$x=7 x=-\frac{4}{7}$$

Steps for Completing the Square

Subtract 45x from both sides.

$$7x^{2}-45x=28$$

Divide both sides by 7.

$$\frac{7x^{2}-45x}{7}=\frac{28}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{45}{7}x=\frac{28}{7}$$

Divide 28 by 7.

$$x^{2}-\frac{45}{7}x=4$$

Divide -\frac{45}{7}, the coefficient of the x term, by 2 to get -\frac{45}{14}. Then add the square of -\frac{45}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{45}{7}x+\left(-\frac{45}{14}\right)^{2}=4+\left(-\frac{45}{14}\right)^{2}$$

Square -\frac{45}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{45}{7}x+\frac{2025}{196}=4+\frac{2025}{196}$$

Add 4 to \frac{2025}{196}.

$$x^{2}-\frac{45}{7}x+\frac{2025}{196}=\frac{2809}{196}$$

Factor x^{2}-\frac{45}{7}x+\frac{2025}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{45}{14}\right)^{2}=\frac{2809}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{45}{14}\right)^{2}}=\sqrt{\frac{2809}{196}}$$

Simplify.

$$x-\frac{45}{14}=\frac{53}{14} x-\frac{45}{14}=-\frac{53}{14}$$

Add \frac{45}{14} to both sides of the equation.

$$x=7 x=-\frac{4}{7}$$