Solve 0=4x^2+5x-21

$$0 = 4 x ^ { 2 } + 5 x - 21$$

Solve for x

$$x=-3$$

$$x = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+5x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=5 ab=4\left(-21\right)=-84$$

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 -84.

$$-1,84 -2,42 -3,28 -4,21 -6,14 -7,12$$

Calculate the sum for each pair.

$$-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5$$

The solution is the pair that gives sum 5.

$$a=-7 b=12$$

Rewrite 4x^{2}+5x-21 as \left(4x^{2}-7x\right)+\left(12x-21\right).

$$\left(4x^{2}-7x\right)+\left(12x-21\right)$$

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

$$x\left(4x-7\right)+3\left(4x-7\right)$$

Factor out common term 4x-7 by using distributive property.

$$\left(4x-7\right)\left(x+3\right)$$

To find equation solutions, solve 4x-7=0 and x+3=0.

$$x=\frac{7}{4} x=-3$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+5x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-21\right)}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-21\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\left(-21\right)}}{2\times 4}$$

Multiply -16 times -21.

$$x=\frac{-5±\sqrt{25+336}}{2\times 4}$$

Add 25 to 336.

$$x=\frac{-5±\sqrt{361}}{2\times 4}$$

Take the square root of 361.

$$x=\frac{-5±19}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±19}{8}$$

Now solve the equation x=\frac{-5±19}{8} when ± is plus. Add -5 to 19.

$$x=\frac{14}{8}$$

Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{4}$$

Now solve the equation x=\frac{-5±19}{8} when ± is minus. Subtract 19 from -5.

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

Divide -24 by 8.

$$x=-3$$

The equation is now solved.

$$x=\frac{7}{4} x=-3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+5x-21=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$4x^{2}+5x=21$$

Divide both sides by 4.

$$\frac{4x^{2}+5x}{4}=\frac{21}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{5}{4}x=\frac{21}{4}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{21}{4}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{21}{4}+\frac{25}{64}$$

Add \frac{21}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{361}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{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}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{361}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{361}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{19}{8} x+\frac{5}{8}=-\frac{19}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{7}{4} x=-3$$