Solve 5x^2=8x-3

$$5 { x }^{ 2 } =8x-3$$

Solve for x

$$x=1$$

$$x=\frac{3}{5}=0.6$$

Steps Using Factoring By Grouping

Subtract 8x from both sides.

$$5x^{2}-8x=-3$$

Add 3 to both sides.

$$5x^{2}-8x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=5\times 3=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite 5x^{2}-8x+3 as \left(5x^{2}-5x\right)+\left(-3x+3\right).

$$\left(5x^{2}-5x\right)+\left(-3x+3\right)$$

Factor out 5x in the first and -3 in the second group.

$$5x\left(x-1\right)-3\left(x-1\right)$$

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

$$\left(x-1\right)\left(5x-3\right)$$

To find equation solutions, solve x-1=0 and 5x-3=0.

$$x=1 x=\frac{3}{5}$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$5x^{2}-8x=-3$$

Add 3 to both sides.

$$5x^{2}-8x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\times 3}}{2\times 5}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 5}$$

Add 64 to -60.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2\times 5}$$

Take the square root of 4.

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

The opposite of -8 is 8.

$$x=\frac{8±2}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{8±2}{10}$$

Now solve the equation x=\frac{8±2}{10} when ± is plus. Add 8 to 2.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{8±2}{10} when ± is minus. Subtract 2 from 8.

$$x=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{5}$$

The equation is now solved.

$$x=1 x=\frac{3}{5}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$5x^{2}-8x=-3$$

Divide both sides by 5.

$$\frac{5x^{2}-8x}{5}=-\frac{3}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{8}{5}x=-\frac{3}{5}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{4}{5}\right)^{2}$$

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

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=-\frac{3}{5}+\frac{16}{25}$$

Add -\frac{3}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{1}{25}$$

Factor x^{2}-\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{5}\right)^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Simplify.

$$x-\frac{4}{5}=\frac{1}{5} x-\frac{4}{5}=-\frac{1}{5}$$

Add \frac{4}{5} to both sides of the equation.

$$x=1 x=\frac{3}{5}$$