Solve 2x^2-8/5-3x-9/10=x-1

$$\frac{ 2 { x }^{ 2 } -8 }{ 5 } - \frac{ 3x-9 }{ 10 } = x-1$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=3$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 10, the least common multiple of 5,10.

$$2\left(2x^{2}-8\right)-\left(3x-9\right)=10x-10$$

Use the distributive property to multiply 2 by 2x^{2}-8.

$$4x^{2}-16-\left(3x-9\right)=10x-10$$

To find the opposite of 3x-9, find the opposite of each term.

$$4x^{2}-16-3x+9=10x-10$$

Add -16 and 9 to get -7.

$$4x^{2}-7-3x=10x-10$$

Subtract 10x from both sides.

$$4x^{2}-7-3x-10x=-10$$

Combine -3x and -10x to get -13x.

$$4x^{2}-7-13x=-10$$

Add 10 to both sides.

$$4x^{2}-7-13x+10=0$$

Add -7 and 10 to get 3.

$$4x^{2}+3-13x=0$$

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

$$4x^{2}-13x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-13 ab=4\times 3=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-12 b=-1$$

Rewrite 4x^{2}-13x+3 as \left(4x^{2}-12x\right)+\left(-x+3\right).

$$\left(4x^{2}-12x\right)+\left(-x+3\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(x-3\right)-\left(x-3\right)$$

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

$$\left(x-3\right)\left(4x-1\right)$$

To find equation solutions, solve x-3=0 and 4x-1=0.

$$x=3 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 10, the least common multiple of 5,10.

$$2\left(2x^{2}-8\right)-\left(3x-9\right)=10x-10$$

Use the distributive property to multiply 2 by 2x^{2}-8.

$$4x^{2}-16-\left(3x-9\right)=10x-10$$

To find the opposite of 3x-9, find the opposite of each term.

$$4x^{2}-16-3x+9=10x-10$$

Add -16 and 9 to get -7.

$$4x^{2}-7-3x=10x-10$$

Subtract 10x from both sides.

$$4x^{2}-7-3x-10x=-10$$

Combine -3x and -10x to get -13x.

$$4x^{2}-7-13x=-10$$

Add 10 to both sides.

$$4x^{2}-7-13x+10=0$$

Add -7 and 10 to get 3.

$$4x^{2}+3-13x=0$$

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.

$$4x^{2}-13x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -13 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-13\right)±\sqrt{169-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$x=\frac{-\left(-13\right)±\sqrt{169-48}}{2\times 4}$$

Add 169 to -48.

$$x=\frac{-\left(-13\right)±\sqrt{121}}{2\times 4}$$

Take the square root of 121.

$$x=\frac{-\left(-13\right)±11}{2\times 4}$$

The opposite of -13 is 13.

$$x=\frac{13±11}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{13±11}{8}$$

Now solve the equation x=\frac{13±11}{8} when ± is plus. Add 13 to 11.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{13±11}{8} when ± is minus. Subtract 11 from 13.

$$x=\frac{2}{8}$$

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

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

The equation is now solved.

$$x=3 x=\frac{1}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 10, the least common multiple of 5,10.

$$2\left(2x^{2}-8\right)-\left(3x-9\right)=10x-10$$

Use the distributive property to multiply 2 by 2x^{2}-8.

$$4x^{2}-16-\left(3x-9\right)=10x-10$$

To find the opposite of 3x-9, find the opposite of each term.

$$4x^{2}-16-3x+9=10x-10$$

Add -16 and 9 to get -7.

$$4x^{2}-7-3x=10x-10$$

Subtract 10x from both sides.

$$4x^{2}-7-3x-10x=-10$$

Combine -3x and -10x to get -13x.

$$4x^{2}-7-13x=-10$$

Add 7 to both sides.

$$4x^{2}-13x=-10+7$$

Add -10 and 7 to get -3.

$$4x^{2}-13x=-3$$

Divide both sides by 4.

$$\frac{4x^{2}-13x}{4}=-\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{13}{4}x=-\frac{3}{4}$$

Divide -\frac{13}{4}, the coefficient of the x term, by 2 to get -\frac{13}{8}. Then add the square of -\frac{13}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{4}x+\left(-\frac{13}{8}\right)^{2}=-\frac{3}{4}+\left(-\frac{13}{8}\right)^{2}$$

Square -\frac{13}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{4}x+\frac{169}{64}=-\frac{3}{4}+\frac{169}{64}$$

Add -\frac{3}{4} to \frac{169}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{4}x+\frac{169}{64}=\frac{121}{64}$$

Factor x^{2}-\frac{13}{4}x+\frac{169}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x-\frac{13}{8}=\frac{11}{8} x-\frac{13}{8}=-\frac{11}{8}$$

Add \frac{13}{8} to both sides of the equation.

$$x=3 x=\frac{1}{4}$$