Solve 6x^2-5left(x-1right)=x*(x+1)+4

$$6 { x }^{ 2 } -5 \left( x-1 \right) = x \times (x+1)+4$$

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply -5 by x-1.

$$6x^{2}-5x+5=x\left(x+1\right)+4$$

Use the distributive property to multiply x by x+1.

$$6x^{2}-5x+5=x^{2}+x+4$$

Subtract x^{2} from both sides.

$$6x^{2}-5x+5-x^{2}=x+4$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-5x+5=x+4$$

Subtract x from both sides.

$$5x^{2}-5x+5-x=4$$

Combine -5x and -x to get -6x.

$$5x^{2}-6x+5=4$$

Subtract 4 from both sides.

$$5x^{2}-6x+5-4=0$$

Subtract 4 from 5 to get 1.

$$5x^{2}-6x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-6 ab=5\times 1=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite 5x^{2}-6x+1 as \left(5x^{2}-5x\right)+\left(-x+1\right).

$$\left(5x^{2}-5x\right)+\left(-x+1\right)$$

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

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

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

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

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

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

Steps Using the Quadratic Formula

Use the distributive property to multiply -5 by x-1.

$$6x^{2}-5x+5=x\left(x+1\right)+4$$

Use the distributive property to multiply x by x+1.

$$6x^{2}-5x+5=x^{2}+x+4$$

Subtract x^{2} from both sides.

$$6x^{2}-5x+5-x^{2}=x+4$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-5x+5=x+4$$

Subtract x from both sides.

$$5x^{2}-5x+5-x=4$$

Combine -5x and -x to get -6x.

$$5x^{2}-6x+5=4$$

Subtract 4 from both sides.

$$5x^{2}-6x+5-4=0$$

Subtract 4 from 5 to get 1.

$$5x^{2}-6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2\times 5}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2\times 5}$$

Add 36 to -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2\times 5}$$

Take the square root of 16.

$$x=\frac{-\left(-6\right)±4}{2\times 5}$$

The opposite of -6 is 6.

$$x=\frac{6±4}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{6±4}{10}$$

Now solve the equation x=\frac{6±4}{10} when ± is plus. Add 6 to 4.

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

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{6±4}{10} when ± is minus. Subtract 4 from 6.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply -5 by x-1.

$$6x^{2}-5x+5=x\left(x+1\right)+4$$

Use the distributive property to multiply x by x+1.

$$6x^{2}-5x+5=x^{2}+x+4$$

Subtract x^{2} from both sides.

$$6x^{2}-5x+5-x^{2}=x+4$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-5x+5=x+4$$

Subtract x from both sides.

$$5x^{2}-5x+5-x=4$$

Combine -5x and -x to get -6x.

$$5x^{2}-6x+5=4$$

Subtract 5 from both sides.

$$5x^{2}-6x=4-5$$

Subtract 5 from 4 to get -1.

$$5x^{2}-6x=-1$$

Divide both sides by 5.

$$\frac{5x^{2}-6x}{5}=-\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{6}{5}x=-\frac{1}{5}$$

Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=-\frac{1}{5}+\left(-\frac{3}{5}\right)^{2}$$

Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=-\frac{1}{5}+\frac{9}{25}$$

Add -\frac{1}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{6}{5}x+\frac{9}{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}.

$$\left(x-\frac{3}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

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

Add \frac{3}{5} to both sides of the equation.

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