Solve -6y^2+11y+10

$$- 6 y ^ { 2 } + 11 y + 10$$

Factor

$$-\left(2y-5\right)\left(3y+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6y^{2}+ay+by+10. To find a and b, set up a system to be solved.

$$a+b=11 ab=-6\times 10=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 11.

$$a=15 b=-4$$

Rewrite -6y^{2}+11y+10 as \left(-6y^{2}+15y\right)+\left(-4y+10\right).

$$\left(-6y^{2}+15y\right)+\left(-4y+10\right)$$

Factor out -3y in the first and -2 in the second group.

$$-3y\left(2y-5\right)-2\left(2y-5\right)$$

Factor out common term 2y-5 by using distributive property.

$$\left(2y-5\right)\left(-3y-2\right)$$

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.

$$-6y^{2}+11y+10=0$$

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.

$$y=\frac{-11±\sqrt{11^{2}-4\left(-6\right)\times 10}}{2\left(-6\right)}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\left(-6\right)\times 10}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$y=\frac{-11±\sqrt{121+24\times 10}}{2\left(-6\right)}$$

Multiply 24 times 10.

$$y=\frac{-11±\sqrt{121+240}}{2\left(-6\right)}$$

Add 121 to 240.

$$y=\frac{-11±\sqrt{361}}{2\left(-6\right)}$$

Take the square root of 361.

$$y=\frac{-11±19}{2\left(-6\right)}$$

Multiply 2 times -6.

$$y=\frac{-11±19}{-12}$$

Now solve the equation y=\frac{-11±19}{-12} when ± is plus. Add -11 to 19.

$$y=\frac{8}{-12}$$

Reduce the fraction \frac{8}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation y=\frac{-11±19}{-12} when ± is minus. Subtract 19 from -11.

$$y=-\frac{30}{-12}$$

Reduce the fraction \frac{-30}{-12} to lowest terms by extracting and canceling out 6.

$$y=\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3} for x_{1} and \frac{5}{2} for x_{2}.

$$-6y^{2}+11y+10=-6\left(y-\left(-\frac{2}{3}\right)\right)\left(y-\frac{5}{2}\right)$$

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

$$-6y^{2}+11y+10=-6\left(y+\frac{2}{3}\right)\left(y-\frac{5}{2}\right)$$

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

$$-6y^{2}+11y+10=-6\times \frac{-3y-2}{-3}\left(y-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6y^{2}+11y+10=-6\times \frac{-3y-2}{-3}\times \frac{-2y+5}{-2}$$

Multiply \frac{-3y-2}{-3} times \frac{-2y+5}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6y^{2}+11y+10=-6\times \frac{\left(-3y-2\right)\left(-2y+5\right)}{-3\left(-2\right)}$$

Multiply -3 times -2.

$$-6y^{2}+11y+10=-6\times \frac{\left(-3y-2\right)\left(-2y+5\right)}{6}$$

Cancel out 6, the greatest common factor in -6 and 6.

$$-6y^{2}+11y+10=-\left(-3y-2\right)\left(-2y+5\right)$$

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.

$$x ^ 2 -\frac{11}{6}x -\frac{5}{3} = 0$$

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

$$r + s = \frac{11}{6} rs = -\frac{5}{3}$$

Two numbers r and s sum up to \frac{11}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{6} = \frac{11}{12}. 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>

$$r = \frac{11}{12} - u s = \frac{11}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(\frac{11}{12} - u) (\frac{11}{12} + u) = -\frac{5}{3}$$

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

$$\frac{121}{144} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{121}{144} on both sides

$$-u^2 = -\frac{5}{3}-\frac{121}{144} = -\frac{361}{144}$$

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

$$u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}$$

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

$$r =\frac{11}{12} - \frac{19}{12} = -0.667 s = \frac{11}{12} + \frac{19}{12} = 2.500$$

Evaluate

$$10+11y-6y^{2}$$