Solve 10x^2+400x-300

$$10 x ^ { 2 } + 400 x - 300$$

Factor

$$10\left(x-\left(-\sqrt{430}-20\right)\right)\left(x-\left(\sqrt{430}-20\right)\right)$$

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.

$$10x^{2}+400x-300=0$$

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.

$$x=\frac{-400±\sqrt{400^{2}-4\times 10\left(-300\right)}}{2\times 10}$$

Square 400.

$$x=\frac{-400±\sqrt{160000-4\times 10\left(-300\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-400±\sqrt{160000-40\left(-300\right)}}{2\times 10}$$

Multiply -40 times -300.

$$x=\frac{-400±\sqrt{160000+12000}}{2\times 10}$$

Add 160000 to 12000.

$$x=\frac{-400±\sqrt{172000}}{2\times 10}$$

Take the square root of 172000.

$$x=\frac{-400±20\sqrt{430}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-400±20\sqrt{430}}{20}$$

Now solve the equation x=\frac{-400±20\sqrt{430}}{20} when ± is plus. Add -400 to 20\sqrt{430}.

$$x=\frac{20\sqrt{430}-400}{20}$$

Divide -400+20\sqrt{430} by 20.

$$x=\sqrt{430}-20$$

Now solve the equation x=\frac{-400±20\sqrt{430}}{20} when ± is minus. Subtract 20\sqrt{430} from -400.

$$x=\frac{-20\sqrt{430}-400}{20}$$

Divide -400-20\sqrt{430} by 20.

$$x=-\sqrt{430}-20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -20+\sqrt{430} for x_{1} and -20-\sqrt{430} for x_{2}.

$$10x^{2}+400x-300=10\left(x-\left(\sqrt{430}-20\right)\right)\left(x-\left(-\sqrt{430}-20\right)\right)$$

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 10

$$x ^ 2 +40x -30 = 0$$

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

$$r + s = -40 rs = -30$$

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

$$r = -20 - u s = -20 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -30

$$(-20 - u) (-20 + u) = -30$$

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

$$400 - u^2 = -30$$

Simplify the expression by subtracting 400 on both sides

$$-u^2 = -30-400 = -430$$

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

$$u^2 = 430 u = \pm\sqrt{430} = \pm \sqrt{430}$$

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

$$r =-20 - \sqrt{430} = -40.736 s = -20 + \sqrt{430} = 0.736$$

Evaluate

$$10\left(x^{2}+40x-30\right)$$