Solve -20x^2+13x+2

$$- 20 x ^ { 2 } + 13 x + 2$$

Factor

$$-20\left(x-\frac{13-\sqrt{329}}{40}\right)\left(x-\frac{\sqrt{329}+13}{40}\right)$$

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.

$$-20x^{2}+13x+2=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-20\right)\times 2}}{2\left(-20\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-20\right)\times 2}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-13±\sqrt{169+80\times 2}}{2\left(-20\right)}$$

Multiply 80 times 2.

$$x=\frac{-13±\sqrt{169+160}}{2\left(-20\right)}$$

Add 169 to 160.

$$x=\frac{-13±\sqrt{329}}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-13±\sqrt{329}}{-40}$$

Now solve the equation x=\frac{-13±\sqrt{329}}{-40} when ± is plus. Add -13 to \sqrt{329}.

$$x=\frac{\sqrt{329}-13}{-40}$$

Divide -13+\sqrt{329} by -40.

$$x=\frac{13-\sqrt{329}}{40}$$

Now solve the equation x=\frac{-13±\sqrt{329}}{-40} when ± is minus. Subtract \sqrt{329} from -13.

$$x=\frac{-\sqrt{329}-13}{-40}$$

Divide -13-\sqrt{329} by -40.

$$x=\frac{\sqrt{329}+13}{40}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13-\sqrt{329}}{40} for x_{1} and \frac{13+\sqrt{329}}{40} for x_{2}.

$$-20x^{2}+13x+2=-20\left(x-\frac{13-\sqrt{329}}{40}\right)\left(x-\frac{\sqrt{329}+13}{40}\right)$$

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.

$$x ^ 2 -\frac{13}{20}x -\frac{1}{10} = 0$$

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

$$r + s = \frac{13}{20} rs = -\frac{1}{10}$$

Two numbers r and s sum up to \frac{13}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{20} = \frac{13}{40}. 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>

$$r = \frac{13}{40} - u s = \frac{13}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{10}

$$(\frac{13}{40} - u) (\frac{13}{40} + u) = -\frac{1}{10}$$

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

$$\frac{169}{1600} - u^2 = -\frac{1}{10}$$

Simplify the expression by subtracting \frac{169}{1600} on both sides

$$-u^2 = -\frac{1}{10}-\frac{169}{1600} = -\frac{329}{1600}$$

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

$$u^2 = \frac{329}{1600} u = \pm\sqrt{\frac{329}{1600}} = \pm \frac{\sqrt{329}}{40}$$

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

$$r =\frac{13}{40} - \frac{\sqrt{329}}{40} = -0.128 s = \frac{13}{40} + \frac{\sqrt{329}}{40} = 0.778$$

Evaluate

$$2+13x-20x^{2}$$