Solve -10x^2+25x+60

$$- 10 x ^ { 2 } + 25 x + 60$$

Factor

$$5\left(4-x\right)\left(2x+3\right)$$

Solution Steps

Factor out 5.

$$5\left(-2x^{2}+5x+12\right)$$

Consider -2x^{2}+5x+12. Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=5 ab=-2\times 12=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=8 b=-3$$

Rewrite -2x^{2}+5x+12 as \left(-2x^{2}+8x\right)+\left(-3x+12\right).

$$\left(-2x^{2}+8x\right)+\left(-3x+12\right)$$

Factor out 2x in the first and 3 in the second group.

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

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

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

Rewrite the complete factored expression.

$$5\left(-x+4\right)\left(2x+3\right)$$

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.

$$-10x^{2}+25x+60=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-10\right)\times 60}}{2\left(-10\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-10\right)\times 60}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-25±\sqrt{625+40\times 60}}{2\left(-10\right)}$$

Multiply 40 times 60.

$$x=\frac{-25±\sqrt{625+2400}}{2\left(-10\right)}$$

Add 625 to 2400.

$$x=\frac{-25±\sqrt{3025}}{2\left(-10\right)}$$

Take the square root of 3025.

$$x=\frac{-25±55}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-25±55}{-20}$$

Now solve the equation x=\frac{-25±55}{-20} when ± is plus. Add -25 to 55.

$$x=\frac{30}{-20}$$

Reduce the fraction \frac{30}{-20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{-25±55}{-20} when ± is minus. Subtract 55 from -25.

$$x=-\frac{80}{-20}$$

Divide -80 by -20.

$$x=4$$

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

$$-10x^{2}+25x+60=-10\left(x-\left(-\frac{3}{2}\right)\right)\left(x-4\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-10x^{2}+25x+60=-10\left(x+\frac{3}{2}\right)\left(x-4\right)$$

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

$$-10x^{2}+25x+60=-10\times \frac{-2x-3}{-2}\left(x-4\right)$$

Cancel out 2, the greatest common factor in -10 and 2.

$$-10x^{2}+25x+60=5\left(-2x-3\right)\left(x-4\right)$$

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.

$$x ^ 2 -\frac{5}{2}x -6 = 0$$

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

$$r + s = \frac{5}{2} rs = -6$$

Two numbers r and s sum up to \frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{2} = \frac{5}{4}. 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>

$$r = \frac{5}{4} - u s = \frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -6

$$(\frac{5}{4} - u) (\frac{5}{4} + u) = -6$$

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

$$\frac{25}{16} - u^2 = -6$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -6-\frac{25}{16} = -\frac{121}{16}$$

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

$$u^2 = \frac{121}{16} u = \pm\sqrt{\frac{121}{16}} = \pm \frac{11}{4}$$

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

$$r =\frac{5}{4} - \frac{11}{4} = -1.500 s = \frac{5}{4} + \frac{11}{4} = 4$$

Evaluate

$$60+25x-10x^{2}$$