Solve 5w^2-40w-50

$$5 w ^ { 2 } - 40 w - 50$$

Factor

$$5\left(w-\left(4-\sqrt{26}\right)\right)\left(w-\left(\sqrt{26}+4\right)\right)$$

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.

$$5w^{2}-40w-50=0$$

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.

$$w=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-50\right)}}{2\times 5}$$

Square -40.

$$w=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-50\right)}}{2\times 5}$$

Multiply -4 times 5.

$$w=\frac{-\left(-40\right)±\sqrt{1600-20\left(-50\right)}}{2\times 5}$$

Multiply -20 times -50.

$$w=\frac{-\left(-40\right)±\sqrt{1600+1000}}{2\times 5}$$

Add 1600 to 1000.

$$w=\frac{-\left(-40\right)±\sqrt{2600}}{2\times 5}$$

Take the square root of 2600.

$$w=\frac{-\left(-40\right)±10\sqrt{26}}{2\times 5}$$

The opposite of -40 is 40.

$$w=\frac{40±10\sqrt{26}}{2\times 5}$$

Multiply 2 times 5.

$$w=\frac{40±10\sqrt{26}}{10}$$

Now solve the equation w=\frac{40±10\sqrt{26}}{10} when ± is plus. Add 40 to 10\sqrt{26}.

$$w=\frac{10\sqrt{26}+40}{10}$$

Divide 40+10\sqrt{26} by 10.

$$w=\sqrt{26}+4$$

Now solve the equation w=\frac{40±10\sqrt{26}}{10} when ± is minus. Subtract 10\sqrt{26} from 40.

$$w=\frac{40-10\sqrt{26}}{10}$$

Divide 40-10\sqrt{26} by 10.

$$w=4-\sqrt{26}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4+\sqrt{26} for x_{1} and 4-\sqrt{26} for x_{2}.

$$5w^{2}-40w-50=5\left(w-\left(\sqrt{26}+4\right)\right)\left(w-\left(4-\sqrt{26}\right)\right)$$

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 5

$$x ^ 2 -8x -10 = 0$$

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

$$r + s = 8 rs = -10$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(4 - u) (4 + u) = -10$$

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

$$16 - u^2 = -10$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -10-16 = -26$$

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

$$u^2 = 26 u = \pm\sqrt{26} = \pm \sqrt{26}$$

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

$$r =4 - \sqrt{26} = -1.099 s = 4 + \sqrt{26} = 9.099$$

Evaluate

$$5\left(w^{2}-8w-10\right)$$