Solve -14x^2+29x-12

$$- 14 x ^ { 2 } + 29 x - 12$$

Factor

$$\left(4-7x\right)\left(2x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -14x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=29 ab=-14\left(-12\right)=168$$

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 168.

$$1,168 2,84 3,56 4,42 6,28 7,24 8,21 12,14$$

Calculate the sum for each pair.

$$1+168=169 2+84=86 3+56=59 4+42=46 6+28=34 7+24=31 8+21=29 12+14=26$$

The solution is the pair that gives sum 29.

$$a=21 b=8$$

Rewrite -14x^{2}+29x-12 as \left(-14x^{2}+21x\right)+\left(8x-12\right).

$$\left(-14x^{2}+21x\right)+\left(8x-12\right)$$

Factor out -7x in the first and 4 in the second group.

$$-7x\left(2x-3\right)+4\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-7x+4\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-14x^{2}+29x-12=0$$

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.

$$x=\frac{-29±\sqrt{29^{2}-4\left(-14\right)\left(-12\right)}}{2\left(-14\right)}$$

Square 29.

$$x=\frac{-29±\sqrt{841-4\left(-14\right)\left(-12\right)}}{2\left(-14\right)}$$

Multiply -4 times -14.

$$x=\frac{-29±\sqrt{841+56\left(-12\right)}}{2\left(-14\right)}$$

Multiply 56 times -12.

$$x=\frac{-29±\sqrt{841-672}}{2\left(-14\right)}$$

Add 841 to -672.

$$x=\frac{-29±\sqrt{169}}{2\left(-14\right)}$$

Take the square root of 169.

$$x=\frac{-29±13}{2\left(-14\right)}$$

Multiply 2 times -14.

$$x=\frac{-29±13}{-28}$$

Now solve the equation x=\frac{-29±13}{-28} when ± is plus. Add -29 to 13.

$$x=-\frac{16}{-28}$$

Reduce the fraction \frac{-16}{-28} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-29±13}{-28} when ± is minus. Subtract 13 from -29.

$$x=-\frac{42}{-28}$$

Reduce the fraction \frac{-42}{-28} to lowest terms by extracting and canceling out 14.

$$x=\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{7} for x_{1} and \frac{3}{2} for x_{2}.

$$-14x^{2}+29x-12=-14\left(x-\frac{4}{7}\right)\left(x-\frac{3}{2}\right)$$

Subtract \frac{4}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-14x^{2}+29x-12=-14\times \frac{-7x+4}{-7}\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-14x^{2}+29x-12=-14\times \frac{-7x+4}{-7}\times \frac{-2x+3}{-2}$$

Multiply \frac{-7x+4}{-7} times \frac{-2x+3}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-14x^{2}+29x-12=-14\times \frac{\left(-7x+4\right)\left(-2x+3\right)}{-7\left(-2\right)}$$

Multiply -7 times -2.

$$-14x^{2}+29x-12=-14\times \frac{\left(-7x+4\right)\left(-2x+3\right)}{14}$$

Cancel out 14, the greatest common factor in -14 and 14.

$$-14x^{2}+29x-12=-\left(-7x+4\right)\left(-2x+3\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -\frac{29}{14}x +\frac{6}{7} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{29}{14} rs = \frac{6}{7}$$

Two numbers r and s sum up to \frac{29}{14} exactly when the average of the two numbers is \frac{1}{2}*\frac{29}{14} = \frac{29}{28}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{29}{28} - u s = \frac{29}{28} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{6}{7}

$$(\frac{29}{28} - u) (\frac{29}{28} + u) = \frac{6}{7}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{841}{784} - u^2 = \frac{6}{7}$$

Simplify the expression by subtracting \frac{841}{784} on both sides

$$-u^2 = \frac{6}{7}-\frac{841}{784} = -\frac{169}{784}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{169}{784} u = \pm\sqrt{\frac{169}{784}} = \pm \frac{13}{28}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{29}{28} - \frac{13}{28} = 0.571 s = \frac{29}{28} + \frac{13}{28} = 1.500$$

Evaluate

$$\left(4-7x\right)\left(2x-3\right)$$