Solve 5x^2=3x+14

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

Solve for x

$$x = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 3x from both sides.

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

Subtract 14 from both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=-3 ab=5\left(-14\right)=-70$$

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 -70.

$$1,-70 2,-35 5,-14 7,-10$$

Calculate the sum for each pair.

$$1-70=-69 2-35=-33 5-14=-9 7-10=-3$$

The solution is the pair that gives sum -3.

$$a=-10 b=7$$

Rewrite 5x^{2}-3x-14 as \left(5x^{2}-10x\right)+\left(7x-14\right).

$$\left(5x^{2}-10x\right)+\left(7x-14\right)$$

Factor out 5x in the first and 7 in the second group.

$$5x\left(x-2\right)+7\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(5x+7\right)$$

To find equation solutions, solve x-2=0 and 5x+7=0.

$$x=2 x=-\frac{7}{5}$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

Subtract 14 from both sides.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 5\left(-14\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-3\right)±\sqrt{9-20\left(-14\right)}}{2\times 5}$$

Multiply -20 times -14.

$$x=\frac{-\left(-3\right)±\sqrt{9+280}}{2\times 5}$$

Add 9 to 280.

$$x=\frac{-\left(-3\right)±\sqrt{289}}{2\times 5}$$

Take the square root of 289.

$$x=\frac{-\left(-3\right)±17}{2\times 5}$$

The opposite of -3 is 3.

$$x=\frac{3±17}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{3±17}{10}$$

Now solve the equation x=\frac{3±17}{10} when ± is plus. Add 3 to 17.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{3±17}{10} when ± is minus. Subtract 17 from 3.

$$x=-\frac{14}{10}$$

Reduce the fraction \frac{-14}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{5}$$

The equation is now solved.

$$x=2 x=-\frac{7}{5}$$

Steps for Completing the Square

Subtract 3x from both sides.

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

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\frac{14}{5}+\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{14}{5}+\frac{9}{100}$$

Add \frac{14}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{289}{100}$$

Factor x^{2}-\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{289}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{289}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{17}{10} x-\frac{3}{10}=-\frac{17}{10}$$

Add \frac{3}{10} to both sides of the equation.

$$x=2 x=-\frac{7}{5}$$