Solve x^2-7x^2+3/10-x^2-3x/2=2x

$$x ^ { 2 } - \frac { 7 x ^ { 2 } + 3 } { 10 } - \frac { x ^ { 2 } - 3 x } { 2 } = 2 x$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=-1$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 10, the least common multiple of 10,2.

$$10x^{2}-\left(7x^{2}+3\right)-5\left(x^{2}-3x\right)=20x$$

To find the opposite of 7x^{2}+3, find the opposite of each term.

$$10x^{2}-7x^{2}-3-5\left(x^{2}-3x\right)=20x$$

Combine 10x^{2} and -7x^{2} to get 3x^{2}.

$$3x^{2}-3-5\left(x^{2}-3x\right)=20x$$

Use the distributive property to multiply -5 by x^{2}-3x.

$$3x^{2}-3-5x^{2}+15x=20x$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-3+15x=20x$$

Subtract 20x from both sides.

$$-2x^{2}-3+15x-20x=0$$

Combine 15x and -20x to get -5x.

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

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

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

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

$$a+b=-5 ab=-2\left(-3\right)=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-2 b=-3$$

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

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

Factor out 2x in the first and 3 in the second group.

$$2x\left(-x-1\right)+3\left(-x-1\right)$$

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

$$\left(-x-1\right)\left(2x+3\right)$$

To find equation solutions, solve -x-1=0 and 2x+3=0.

$$x=-1 x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 10, the least common multiple of 10,2.

$$10x^{2}-\left(7x^{2}+3\right)-5\left(x^{2}-3x\right)=20x$$

To find the opposite of 7x^{2}+3, find the opposite of each term.

$$10x^{2}-7x^{2}-3-5\left(x^{2}-3x\right)=20x$$

Combine 10x^{2} and -7x^{2} to get 3x^{2}.

$$3x^{2}-3-5\left(x^{2}-3x\right)=20x$$

Use the distributive property to multiply -5 by x^{2}-3x.

$$3x^{2}-3-5x^{2}+15x=20x$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-3+15x=20x$$

Subtract 20x from both sides.

$$-2x^{2}-3+15x-20x=0$$

Combine 15x and -20x to get -5x.

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

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.

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

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

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

Square -5.

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

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2\left(-2\right)}$$

Add 25 to -24.

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

Take the square root of 1.

$$x=\frac{-\left(-5\right)±1}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±1}{-4}$$

Now solve the equation x=\frac{5±1}{-4} when ± is plus. Add 5 to 1.

$$x=\frac{6}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{5±1}{-4} when ± is minus. Subtract 1 from 5.

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

Divide 4 by -4.

$$x=-1$$

The equation is now solved.

$$x=-\frac{3}{2} x=-1$$

Steps for Completing the Square

Multiply both sides of the equation by 10, the least common multiple of 10,2.

$$10x^{2}-\left(7x^{2}+3\right)-5\left(x^{2}-3x\right)=20x$$

To find the opposite of 7x^{2}+3, find the opposite of each term.

$$10x^{2}-7x^{2}-3-5\left(x^{2}-3x\right)=20x$$

Combine 10x^{2} and -7x^{2} to get 3x^{2}.

$$3x^{2}-3-5\left(x^{2}-3x\right)=20x$$

Use the distributive property to multiply -5 by x^{2}-3x.

$$3x^{2}-3-5x^{2}+15x=20x$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-3+15x=20x$$

Subtract 20x from both sides.

$$-2x^{2}-3+15x-20x=0$$

Combine 15x and -20x to get -5x.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -5 by -2.

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

Divide 3 by -2.

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

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

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

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

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

Add -\frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$\left(x+\frac{5}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

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

Subtract \frac{5}{4} from both sides of the equation.

$$x=-1 x=-\frac{3}{2}$$