Solve 6x^2-6=35x

$$6 x ^ { 2 } - 6 = 35 x$$

Solve for x

$$x=-\frac{1}{6}\approx -0.166666667$$

$$x=6$$

Steps Using Factoring By Grouping

Subtract 35x from both sides.

$$6x^{2}-6-35x=0$$

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

$$6x^{2}-35x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-35 ab=6\left(-6\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -35.

$$a=-36 b=1$$

Rewrite 6x^{2}-35x-6 as \left(6x^{2}-36x\right)+\left(x-6\right).

$$\left(6x^{2}-36x\right)+\left(x-6\right)$$

Factor out 6x in 6x^{2}-36x.

$$6x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(6x+1\right)$$

To find equation solutions, solve x-6=0 and 6x+1=0.

$$x=6 x=-\frac{1}{6}$$

Steps Using the Quadratic Formula

Subtract 35x from both sides.

$$6x^{2}-6-35x=0$$

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.

$$6x^{2}-35x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -35 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 6\left(-6\right)}}{2\times 6}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 6\left(-6\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-35\right)±\sqrt{1225-24\left(-6\right)}}{2\times 6}$$

Multiply -24 times -6.

$$x=\frac{-\left(-35\right)±\sqrt{1225+144}}{2\times 6}$$

Add 1225 to 144.

$$x=\frac{-\left(-35\right)±\sqrt{1369}}{2\times 6}$$

Take the square root of 1369.

$$x=\frac{-\left(-35\right)±37}{2\times 6}$$

The opposite of -35 is 35.

$$x=\frac{35±37}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{35±37}{12}$$

Now solve the equation x=\frac{35±37}{12} when ± is plus. Add 35 to 37.

$$x=\frac{72}{12}$$

Divide 72 by 12.

$$x=6$$

Now solve the equation x=\frac{35±37}{12} when ± is minus. Subtract 37 from 35.

$$x=-\frac{2}{12}$$

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{6}$$

The equation is now solved.

$$x=6 x=-\frac{1}{6}$$

Steps for Completing the Square

Subtract 35x from both sides.

$$6x^{2}-6-35x=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$6x^{2}-35x=6$$

Divide both sides by 6.

$$\frac{6x^{2}-35x}{6}=\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{35}{6}x=\frac{6}{6}$$

Divide 6 by 6.

$$x^{2}-\frac{35}{6}x=1$$

Divide -\frac{35}{6}, the coefficient of the x term, by 2 to get -\frac{35}{12}. Then add the square of -\frac{35}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{6}x+\left(-\frac{35}{12}\right)^{2}=1+\left(-\frac{35}{12}\right)^{2}$$

Square -\frac{35}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{6}x+\frac{1225}{144}=1+\frac{1225}{144}$$

Add 1 to \frac{1225}{144}.

$$x^{2}-\frac{35}{6}x+\frac{1225}{144}=\frac{1369}{144}$$

Factor x^{2}-\frac{35}{6}x+\frac{1225}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{35}{12}\right)^{2}=\frac{1369}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{12}\right)^{2}}=\sqrt{\frac{1369}{144}}$$

Simplify.

$$x-\frac{35}{12}=\frac{37}{12} x-\frac{35}{12}=-\frac{37}{12}$$

Add \frac{35}{12} to both sides of the equation.

$$x=6 x=-\frac{1}{6}$$