Solve 5x^2-10x+3=6x

$$5 x ^ { 2 } - 10 x + 3 = 6 x$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 6x from both sides.

$$5x^{2}-10x+3-6x=0$$

Combine -10x and -6x to get -16x.

$$5x^{2}-16x+3=0$$

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.

$$a+b=-16 ab=5\times 3=15$$

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 -16.

$$a=-15 b=-1$$

Rewrite 5x^{2}-16x+3 as \left(5x^{2}-15x\right)+\left(-x+3\right).

$$\left(5x^{2}-15x\right)+\left(-x+3\right)$$

Factor out 5x in the first and -1 in the second group.

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

Factor out common term x-3 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 6x from both sides.

$$5x^{2}-10x+3-6x=0$$

Combine -10x and -6x to get -16x.

$$5x^{2}-16x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -16 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 3}}{2\times 5}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-16\right)±\sqrt{256-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$x=\frac{-\left(-16\right)±\sqrt{256-60}}{2\times 5}$$

Add 256 to -60.

$$x=\frac{-\left(-16\right)±\sqrt{196}}{2\times 5}$$

Take the square root of 196.

$$x=\frac{-\left(-16\right)±14}{2\times 5}$$

The opposite of -16 is 16.

$$x=\frac{16±14}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{16±14}{10}$$

Now solve the equation x=\frac{16±14}{10} when ± is plus. Add 16 to 14.

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{16±14}{10} when ± is minus. Subtract 14 from 16.

$$x=\frac{2}{10}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 6x from both sides.

$$5x^{2}-10x+3-6x=0$$

Combine -10x and -6x to get -16x.

$$5x^{2}-16x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-16x=-3$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{5}x+\left(-\frac{8}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{8}{5}\right)^{2}$$

Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{5}x+\frac{64}{25}=-\frac{3}{5}+\frac{64}{25}$$

Add -\frac{3}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{16}{5}x+\frac{64}{25}=\frac{49}{25}$$

Factor x^{2}-\frac{16}{5}x+\frac{64}{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}.

$$\left(x-\frac{8}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$x-\frac{8}{5}=\frac{7}{5} x-\frac{8}{5}=-\frac{7}{5}$$

Add \frac{8}{5} to both sides of the equation.

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