Solve 2x^2-29x-36=

$$2 x ^ { 2 } - 29 x - 36 =$$

Factor

$$2\left(x-\frac{29-\sqrt{1129}}{4}\right)\left(x-\frac{\sqrt{1129}+29}{4}\right)$$

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.

$$2x^{2}-29x-36=0$$

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.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 2\left(-36\right)}}{2\times 2}$$

Square -29.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\times 2\left(-36\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-29\right)±\sqrt{841-8\left(-36\right)}}{2\times 2}$$

Multiply -8 times -36.

$$x=\frac{-\left(-29\right)±\sqrt{841+288}}{2\times 2}$$

Add 841 to 288.

$$x=\frac{-\left(-29\right)±\sqrt{1129}}{2\times 2}$$

The opposite of -29 is 29.

$$x=\frac{29±\sqrt{1129}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{29±\sqrt{1129}}{4}$$

Now solve the equation x=\frac{29±\sqrt{1129}}{4} when ± is plus. Add 29 to \sqrt{1129}.

$$x=\frac{\sqrt{1129}+29}{4}$$

Now solve the equation x=\frac{29±\sqrt{1129}}{4} when ± is minus. Subtract \sqrt{1129} from 29.

$$x=\frac{29-\sqrt{1129}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{29+\sqrt{1129}}{4} for x_{1} and \frac{29-\sqrt{1129}}{4} for x_{2}.

$$2x^{2}-29x-36=2\left(x-\frac{\sqrt{1129}+29}{4}\right)\left(x-\frac{29-\sqrt{1129}}{4}\right)$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 -\frac{29}{2}x -18 = 0$$

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

$$r + s = \frac{29}{2} rs = -18$$

Two numbers r and s sum up to \frac{29}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{29}{2} = \frac{29}{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>

$$r = \frac{29}{4} - u s = \frac{29}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -18

$$(\frac{29}{4} - u) (\frac{29}{4} + u) = -18$$

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

$$\frac{841}{16} - u^2 = -18$$

Simplify the expression by subtracting \frac{841}{16} on both sides

$$-u^2 = -18-\frac{841}{16} = -\frac{1129}{16}$$

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

$$u^2 = \frac{1129}{16} u = \pm\sqrt{\frac{1129}{16}} = \pm \frac{\sqrt{1129}}{4}$$

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

$$r =\frac{29}{4} - \frac{\sqrt{1129}}{4} = -1.150 s = \frac{29}{4} + \frac{\sqrt{1129}}{4} = 15.650$$

Evaluate

$$2x^{2}-29x-36$$