Solve -2y^2+11y+3

$$- 2 y ^ { 2 } + 11 y + 3$$

Factor

$$-2\left(y-\frac{11-\sqrt{145}}{4}\right)\left(y-\frac{\sqrt{145}+11}{4}\right)$$

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.

$$-2y^{2}+11y+3=0$$

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.

$$y=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$y=\frac{-11±\sqrt{121+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

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

Add 121 to 24.

$$y=\frac{-11±\sqrt{145}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{-11±\sqrt{145}}{-4}$$

Now solve the equation y=\frac{-11±\sqrt{145}}{-4} when ± is plus. Add -11 to \sqrt{145}.

$$y=\frac{\sqrt{145}-11}{-4}$$

Divide -11+\sqrt{145} by -4.

$$y=\frac{11-\sqrt{145}}{4}$$

Now solve the equation y=\frac{-11±\sqrt{145}}{-4} when ± is minus. Subtract \sqrt{145} from -11.

$$y=\frac{-\sqrt{145}-11}{-4}$$

Divide -11-\sqrt{145} by -4.

$$y=\frac{\sqrt{145}+11}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{11-\sqrt{145}}{4} for x_{1} and \frac{11+\sqrt{145}}{4} for x_{2}.

$$-2y^{2}+11y+3=-2\left(y-\frac{11-\sqrt{145}}{4}\right)\left(y-\frac{\sqrt{145}+11}{4}\right)$$

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.

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

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

$$r + s = \frac{11}{2} rs = -\frac{3}{2}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = -\frac{3}{2}$$

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

$$\frac{121}{16} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = -\frac{3}{2}-\frac{121}{16} = -\frac{145}{16}$$

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

$$u^2 = \frac{145}{16} u = \pm\sqrt{\frac{145}{16}} = \pm \frac{\sqrt{145}}{4}$$

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

$$r =\frac{11}{4} - \frac{\sqrt{145}}{4} = -0.260 s = \frac{11}{4} + \frac{\sqrt{145}}{4} = 5.760$$

Evaluate

$$3+11y-2y^{2}$$