Solve 30x+21x^2=3384

$$30 x + 21 x ^ { 2 } = 3384$$

Solve for x

$$x = -\frac{94}{7} = -13\frac{3}{7} \approx -13.428571429$$

$$x=12$$

Steps Using Factoring By Grouping

Subtract 3384 from both sides.

$$30x+21x^{2}-3384=0$$

Divide both sides by 3.

$$10x+7x^{2}-1128=0$$

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

$$7x^{2}+10x-1128=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-1128. To find a and b, set up a system to be solved.

$$a+b=10 ab=7\left(-1128\right)=-7896$$

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 -7896.

$$-1,7896 -2,3948 -3,2632 -4,1974 -6,1316 -7,1128 -8,987 -12,658 -14,564 -21,376 -24,329 -28,282 -42,188 -47,168 -56,141 -84,94$$

Calculate the sum for each pair.

$$-1+7896=7895 -2+3948=3946 -3+2632=2629 -4+1974=1970 -6+1316=1310 -7+1128=1121 -8+987=979 -12+658=646 -14+564=550 -21+376=355 -24+329=305 -28+282=254 -42+188=146 -47+168=121 -56+141=85 -84+94=10$$

The solution is the pair that gives sum 10.

$$a=-84 b=94$$

Rewrite 7x^{2}+10x-1128 as \left(7x^{2}-84x\right)+\left(94x-1128\right).

$$\left(7x^{2}-84x\right)+\left(94x-1128\right)$$

Factor out 7x in the first and 94 in the second group.

$$7x\left(x-12\right)+94\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(7x+94\right)$$

To find equation solutions, solve x-12=0 and 7x+94=0.

$$x=12 x=-\frac{94}{7}$$

Steps Using the Quadratic Formula

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.

$$21x^{2}+30x=3384$$

Subtract 3384 from both sides of the equation.

$$21x^{2}+30x-3384=3384-3384$$

Subtracting 3384 from itself leaves 0.

$$21x^{2}+30x-3384=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, 30 for b, and -3384 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\times 21\left(-3384\right)}}{2\times 21}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\times 21\left(-3384\right)}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-30±\sqrt{900-84\left(-3384\right)}}{2\times 21}$$

Multiply -84 times -3384.

$$x=\frac{-30±\sqrt{900+284256}}{2\times 21}$$

Add 900 to 284256.

$$x=\frac{-30±\sqrt{285156}}{2\times 21}$$

Take the square root of 285156.

$$x=\frac{-30±534}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-30±534}{42}$$

Now solve the equation x=\frac{-30±534}{42} when ± is plus. Add -30 to 534.

$$x=\frac{504}{42}$$

Divide 504 by 42.

$$x=12$$

Now solve the equation x=\frac{-30±534}{42} when ± is minus. Subtract 534 from -30.

$$x=-\frac{564}{42}$$

Reduce the fraction \frac{-564}{42} to lowest terms by extracting and canceling out 6.

$$x=-\frac{94}{7}$$

The equation is now solved.

$$x=12 x=-\frac{94}{7}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$21x^{2}+30x=3384$$

Divide both sides by 21.

$$\frac{21x^{2}+30x}{21}=\frac{3384}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$x^{2}+\frac{30}{21}x=\frac{3384}{21}$$

Reduce the fraction \frac{30}{21} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{10}{7}x=\frac{3384}{21}$$

Reduce the fraction \frac{3384}{21} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{10}{7}x=\frac{1128}{7}$$

Divide \frac{10}{7}, the coefficient of the x term, by 2 to get \frac{5}{7}. Then add the square of \frac{5}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{7}x+\left(\frac{5}{7}\right)^{2}=\frac{1128}{7}+\left(\frac{5}{7}\right)^{2}$$

Square \frac{5}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{7}x+\frac{25}{49}=\frac{1128}{7}+\frac{25}{49}$$

Add \frac{1128}{7} to \frac{25}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{7}x+\frac{25}{49}=\frac{7921}{49}$$

Factor x^{2}+\frac{10}{7}x+\frac{25}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{7}\right)^{2}=\frac{7921}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{7}\right)^{2}}=\sqrt{\frac{7921}{49}}$$

Simplify.

$$x+\frac{5}{7}=\frac{89}{7} x+\frac{5}{7}=-\frac{89}{7}$$

Subtract \frac{5}{7} from both sides of the equation.

$$x=12 x=-\frac{94}{7}$$