Solve -7R^2+20R-12

$$- 7 R ^ { 2 } + 20 R - 12$$

Factor

$$\left(2-R\right)\left(7R-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -7R^{2}+aR+bR-12. To find a and b, set up a system to be solved.

$$a+b=20 ab=-7\left(-12\right)=84$$

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 84.

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

Calculate the sum for each pair.

$$1+84=85 2+42=44 3+28=31 4+21=25 6+14=20 7+12=19$$

The solution is the pair that gives sum 20.

$$a=14 b=6$$

Rewrite -7R^{2}+20R-12 as \left(-7R^{2}+14R\right)+\left(6R-12\right).

$$\left(-7R^{2}+14R\right)+\left(6R-12\right)$$

Factor out 7R in the first and -6 in the second group.

$$7R\left(-R+2\right)-6\left(-R+2\right)$$

Factor out common term -R+2 by using distributive property.

$$\left(-R+2\right)\left(7R-6\right)$$

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.

$$-7R^{2}+20R-12=0$$

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.

$$R=\frac{-20±\sqrt{20^{2}-4\left(-7\right)\left(-12\right)}}{2\left(-7\right)}$$

Square 20.

$$R=\frac{-20±\sqrt{400-4\left(-7\right)\left(-12\right)}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$R=\frac{-20±\sqrt{400+28\left(-12\right)}}{2\left(-7\right)}$$

Multiply 28 times -12.

$$R=\frac{-20±\sqrt{400-336}}{2\left(-7\right)}$$

Add 400 to -336.

$$R=\frac{-20±\sqrt{64}}{2\left(-7\right)}$$

Take the square root of 64.

$$R=\frac{-20±8}{2\left(-7\right)}$$

Multiply 2 times -7.

$$R=\frac{-20±8}{-14}$$

Now solve the equation R=\frac{-20±8}{-14} when ± is plus. Add -20 to 8.

$$R=-\frac{12}{-14}$$

Reduce the fraction \frac{-12}{-14} to lowest terms by extracting and canceling out 2.

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

Now solve the equation R=\frac{-20±8}{-14} when ± is minus. Subtract 8 from -20.

$$R=-\frac{28}{-14}$$

Divide -28 by -14.

$$R=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{6}{7} for x_{1} and 2 for x_{2}.

$$-7R^{2}+20R-12=-7\left(R-\frac{6}{7}\right)\left(R-2\right)$$

Subtract \frac{6}{7} from R by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-7R^{2}+20R-12=-7\times \frac{-7R+6}{-7}\left(R-2\right)$$

Cancel out 7, the greatest common factor in -7 and 7.

$$-7R^{2}+20R-12=\left(-7R+6\right)\left(R-2\right)$$

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.

$$x ^ 2 -\frac{20}{7}x +\frac{12}{7} = 0$$

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

$$r + s = \frac{20}{7} rs = \frac{12}{7}$$

Two numbers r and s sum up to \frac{20}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{20}{7} = \frac{10}{7}. 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>

$$r = \frac{10}{7} - u s = \frac{10}{7} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{12}{7}

$$(\frac{10}{7} - u) (\frac{10}{7} + u) = \frac{12}{7}$$

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

$$\frac{100}{49} - u^2 = \frac{12}{7}$$

Simplify the expression by subtracting \frac{100}{49} on both sides

$$-u^2 = \frac{12}{7}-\frac{100}{49} = -\frac{16}{49}$$

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

$$u^2 = \frac{16}{49} u = \pm\sqrt{\frac{16}{49}} = \pm \frac{4}{7}$$

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

$$r =\frac{10}{7} - \frac{4}{7} = 0.857 s = \frac{10}{7} + \frac{4}{7} = 2$$

Evaluate

$$\left(2-R\right)\left(7R-6\right)$$