Solve 4x^2+5=-21x

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

Solve for x

$$x=-5$$

$$x=-\frac{1}{4}=-0.25$$

Steps Using Factoring By Grouping

Add 21x to both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=21 ab=4\times 5=20$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 21.

$$a=1 b=20$$

Rewrite 4x^{2}+21x+5 as \left(4x^{2}+x\right)+\left(20x+5\right).

$$\left(4x^{2}+x\right)+\left(20x+5\right)$$

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

$$x\left(4x+1\right)+5\left(4x+1\right)$$

Factor out common term 4x+1 by using distributive property.

$$\left(4x+1\right)\left(x+5\right)$$

To find equation solutions, solve 4x+1=0 and x+5=0.

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

Steps Using the Quadratic Formula

Add 21x to both sides.

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

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.

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

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

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

Square 21.

$$x=\frac{-21±\sqrt{441-4\times 4\times 5}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-21±\sqrt{441-16\times 5}}{2\times 4}$$

Multiply -16 times 5.

$$x=\frac{-21±\sqrt{441-80}}{2\times 4}$$

Add 441 to -80.

$$x=\frac{-21±\sqrt{361}}{2\times 4}$$

Take the square root of 361.

$$x=\frac{-21±19}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-21±19}{8}$$

Now solve the equation x=\frac{-21±19}{8} when ± is plus. Add -21 to 19.

$$x=-\frac{2}{8}$$

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

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

Now solve the equation x=\frac{-21±19}{8} when ± is minus. Subtract 19 from -21.

$$x=-\frac{40}{8}$$

Divide -40 by 8.

$$x=-5$$

The equation is now solved.

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

Steps for Completing the Square

Add 21x to both sides.

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

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide \frac{21}{4}, the coefficient of the x term, by 2 to get \frac{21}{8}. Then add the square of \frac{21}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{21}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{21}{4}x+\frac{441}{64}=-\frac{5}{4}+\frac{441}{64}$$

Add -\frac{5}{4} to \frac{441}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{21}{4}x+\frac{441}{64}=\frac{361}{64}$$

Factor x^{2}+\frac{21}{4}x+\frac{441}{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}.

$$\left(x+\frac{21}{8}\right)^{2}=\frac{361}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{21}{8}\right)^{2}}=\sqrt{\frac{361}{64}}$$

Simplify.

$$x+\frac{21}{8}=\frac{19}{8} x+\frac{21}{8}=-\frac{19}{8}$$

Subtract \frac{21}{8} from both sides of the equation.

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