Solve 2x^2+14x-4=-x^2+3x

$$2 x ^ { 2 } + 14 x - 4 = - x ^ { 2 } + 3 x$$

Solve for x

$$x=-4$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Add x^{2} to both sides.

$$2x^{2}+14x-4+x^{2}=3x$$

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

$$3x^{2}+14x-4=3x$$

Subtract 3x from both sides.

$$3x^{2}+14x-4-3x=0$$

Combine 14x and -3x to get 11x.

$$3x^{2}+11x-4=0$$

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

$$a+b=11 ab=3\left(-4\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

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

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 11.

$$a=-1 b=12$$

Rewrite 3x^{2}+11x-4 as \left(3x^{2}-x\right)+\left(12x-4\right).

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

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

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

Factor out common term 3x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$2x^{2}+14x-4+x^{2}=3x$$

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

$$3x^{2}+14x-4=3x$$

Subtract 3x from both sides.

$$3x^{2}+14x-4-3x=0$$

Combine 14x and -3x to get 11x.

$$3x^{2}+11x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 11 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 3\left(-4\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-11±\sqrt{121-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-11±\sqrt{121+48}}{2\times 3}$$

Add 121 to 48.

$$x=\frac{-11±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-11±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-11±13}{6}$$

Now solve the equation x=\frac{-11±13}{6} when ± is plus. Add -11 to 13.

$$x=\frac{2}{6}$$

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

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

Now solve the equation x=\frac{-11±13}{6} when ± is minus. Subtract 13 from -11.

$$x=-\frac{24}{6}$$

Divide -24 by 6.

$$x=-4$$

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

$$2x^{2}+14x-4+x^{2}=3x$$

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

$$3x^{2}+14x-4=3x$$

Subtract 3x from both sides.

$$3x^{2}+14x-4-3x=0$$

Combine 14x and -3x to get 11x.

$$3x^{2}+11x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$3x^{2}+11x=4$$

Divide both sides by 3.

$$\frac{3x^{2}+11x}{3}=\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{11}{3}x=\frac{4}{3}$$

Divide \frac{11}{3}, the coefficient of the x term, by 2 to get \frac{11}{6}. Then add the square of \frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{3}x+\left(\frac{11}{6}\right)^{2}=\frac{4}{3}+\left(\frac{11}{6}\right)^{2}$$

Square \frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{4}{3}+\frac{121}{36}$$

Add \frac{4}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{169}{36}$$

Factor x^{2}+\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x+\frac{11}{6}=\frac{13}{6} x+\frac{11}{6}=-\frac{13}{6}$$

Subtract \frac{11}{6} from both sides of the equation.

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