Solve -3y^2-4y+12

$$- 3 y ^ { 2 } - 4 y + 12$$

Factor

$$-3\left(y-\frac{-2\sqrt{10}-2}{3}\right)\left(y-\frac{2\sqrt{10}-2}{3}\right)$$

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.

$$-3y^{2}-4y+12=0$$

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.

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-\left(-4\right)±\sqrt{16+12\times 12}}{2\left(-3\right)}$$

Multiply 12 times 12.

$$y=\frac{-\left(-4\right)±\sqrt{16+144}}{2\left(-3\right)}$$

Add 16 to 144.

$$y=\frac{-\left(-4\right)±\sqrt{160}}{2\left(-3\right)}$$

Take the square root of 160.

$$y=\frac{-\left(-4\right)±4\sqrt{10}}{2\left(-3\right)}$$

The opposite of -4 is 4.

$$y=\frac{4±4\sqrt{10}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{4±4\sqrt{10}}{-6}$$

Now solve the equation y=\frac{4±4\sqrt{10}}{-6} when ± is plus. Add 4 to 4\sqrt{10}.

$$y=\frac{4\sqrt{10}+4}{-6}$$

Divide 4+4\sqrt{10} by -6.

$$y=\frac{-2\sqrt{10}-2}{3}$$

Now solve the equation y=\frac{4±4\sqrt{10}}{-6} when ± is minus. Subtract 4\sqrt{10} from 4.

$$y=\frac{4-4\sqrt{10}}{-6}$$

Divide 4-4\sqrt{10} by -6.

$$y=\frac{2\sqrt{10}-2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-2-2\sqrt{10}}{3} for x_{1} and \frac{-2+2\sqrt{10}}{3} for x_{2}.

$$-3y^{2}-4y+12=-3\left(y-\frac{-2\sqrt{10}-2}{3}\right)\left(y-\frac{2\sqrt{10}-2}{3}\right)$$

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.

$$x ^ 2 +\frac{4}{3}x -4 = 0$$

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

$$r + s = -\frac{4}{3} rs = -4$$

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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>

$$r = -\frac{2}{3} - u s = -\frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = -4$$

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

$$\frac{4}{9} - u^2 = -4$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = -4-\frac{4}{9} = -\frac{40}{9}$$

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

$$u^2 = \frac{40}{9} u = \pm\sqrt{\frac{40}{9}} = \pm \frac{\sqrt{40}}{3}$$

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

$$r =-\frac{2}{3} - \frac{\sqrt{40}}{3} = -2.775 s = -\frac{2}{3} + \frac{\sqrt{40}}{3} = 1.442$$

Evaluate

$$12-4y-3y^{2}$$