Solve 3x^2=5+14x

$$3 x ^ { 2 } = 5 + 14 x$$

Solve for x

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

$$x=5$$

Steps Using Factoring By Grouping

Subtract 5 from both sides.

$$3x^{2}-5=14x$$

Subtract 14x from both sides.

$$3x^{2}-5-14x=0$$

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

$$3x^{2}-14x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-14 ab=3\left(-5\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -14.

$$a=-15 b=1$$

Rewrite 3x^{2}-14x-5 as \left(3x^{2}-15x\right)+\left(x-5\right).

$$\left(3x^{2}-15x\right)+\left(x-5\right)$$

Factor out 3x in 3x^{2}-15x.

$$3x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(3x+1\right)$$

To find equation solutions, solve x-5=0 and 3x+1=0.

$$x=5 x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 5 from both sides.

$$3x^{2}-5=14x$$

Subtract 14x from both sides.

$$3x^{2}-5-14x=0$$

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.

$$3x^{2}-14x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -14 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-14\right)±\sqrt{196-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$x=\frac{-\left(-14\right)±\sqrt{196+60}}{2\times 3}$$

Add 196 to 60.

$$x=\frac{-\left(-14\right)±\sqrt{256}}{2\times 3}$$

Take the square root of 256.

$$x=\frac{-\left(-14\right)±16}{2\times 3}$$

The opposite of -14 is 14.

$$x=\frac{14±16}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{14±16}{6}$$

Now solve the equation x=\frac{14±16}{6} when ± is plus. Add 14 to 16.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{14±16}{6} when ± is minus. Subtract 16 from 14.

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

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

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

The equation is now solved.

$$x=5 x=-\frac{1}{3}$$

Steps for Completing the Square

Subtract 14x from both sides.

$$3x^{2}-14x=5$$

Divide both sides by 3.

$$\frac{3x^{2}-14x}{3}=\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{14}{3}x=\frac{5}{3}$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=\frac{5}{3}+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{5}{3}+\frac{49}{9}$$

Add \frac{5}{3} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{7}{3}=\frac{8}{3} x-\frac{7}{3}=-\frac{8}{3}$$

Add \frac{7}{3} to both sides of the equation.

$$x=5 x=-\frac{1}{3}$$