Solve 21x^2+5x-35=3x^2-4x

$$21 x ^ { 2 } + 5 x - 35 = 3 x ^ { 2 } - 4 x$$

Solve for x

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

Steps Using Factoring By Grouping

Subtract 3x^{2} from both sides.

$$21x^{2}+5x-35-3x^{2}=-4x$$

Combine 21x^{2} and -3x^{2} to get 18x^{2}.

$$18x^{2}+5x-35=-4x$$

Add 4x to both sides.

$$18x^{2}+5x-35+4x=0$$

Combine 5x and 4x to get 9x.

$$18x^{2}+9x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 18x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=9 ab=18\left(-35\right)=-630$$

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 -630.

$$-1,630 -2,315 -3,210 -5,126 -6,105 -7,90 -9,70 -10,63 -14,45 -15,42 -18,35 -21,30$$

Calculate the sum for each pair.

$$-1+630=629 -2+315=313 -3+210=207 -5+126=121 -6+105=99 -7+90=83 -9+70=61 -10+63=53 -14+45=31 -15+42=27 -18+35=17 -21+30=9$$

The solution is the pair that gives sum 9.

$$a=-21 b=30$$

Rewrite 18x^{2}+9x-35 as \left(18x^{2}-21x\right)+\left(30x-35\right).

$$\left(18x^{2}-21x\right)+\left(30x-35\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(6x-7\right)+5\left(6x-7\right)$$

Factor out common term 6x-7 by using distributive property.

$$\left(6x-7\right)\left(3x+5\right)$$

To find equation solutions, solve 6x-7=0 and 3x+5=0.

$$x=\frac{7}{6} x=-\frac{5}{3}$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$21x^{2}+5x-35-3x^{2}=-4x$$

Combine 21x^{2} and -3x^{2} to get 18x^{2}.

$$18x^{2}+5x-35=-4x$$

Add 4x to both sides.

$$18x^{2}+5x-35+4x=0$$

Combine 5x and 4x to get 9x.

$$18x^{2}+9x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 9 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 18\left(-35\right)}}{2\times 18}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 18\left(-35\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-9±\sqrt{81-72\left(-35\right)}}{2\times 18}$$

Multiply -72 times -35.

$$x=\frac{-9±\sqrt{81+2520}}{2\times 18}$$

Add 81 to 2520.

$$x=\frac{-9±\sqrt{2601}}{2\times 18}$$

Take the square root of 2601.

$$x=\frac{-9±51}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-9±51}{36}$$

Now solve the equation x=\frac{-9±51}{36} when ± is plus. Add -9 to 51.

$$x=\frac{42}{36}$$

Reduce the fraction \frac{42}{36} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{6}$$

Now solve the equation x=\frac{-9±51}{36} when ± is minus. Subtract 51 from -9.

$$x=-\frac{60}{36}$$

Reduce the fraction \frac{-60}{36} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

$$x=\frac{7}{6} x=-\frac{5}{3}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$21x^{2}+5x-35-3x^{2}=-4x$$

Combine 21x^{2} and -3x^{2} to get 18x^{2}.

$$18x^{2}+5x-35=-4x$$

Add 4x to both sides.

$$18x^{2}+5x-35+4x=0$$

Combine 5x and 4x to get 9x.

$$18x^{2}+9x-35=0$$

Add 35 to both sides. Anything plus zero gives itself.

$$18x^{2}+9x=35$$

Divide both sides by 18.

$$\frac{18x^{2}+9x}{18}=\frac{35}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\frac{9}{18}x=\frac{35}{18}$$

Reduce the fraction \frac{9}{18} to lowest terms by extracting and canceling out 9.

$$x^{2}+\frac{1}{2}x=\frac{35}{18}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{35}{18}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{35}{18}+\frac{1}{16}$$

Add \frac{35}{18} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{289}{144}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$x+\frac{1}{4}=\frac{17}{12} x+\frac{1}{4}=-\frac{17}{12}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{7}{6} x=-\frac{5}{3}$$