Solve 15x^2-525x-4500=0

$$15 x ^ { 2 } - 525 x - 4500 = 0$$

Solve for x

$$x = \frac{5 \sqrt{97} + 35}{2} \approx 42.122144504$$

$$x=\frac{35-5\sqrt{97}}{2}\approx -7.122144504$$

Steps Using the Quadratic Formula

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.

$$15x^{2}-525x-4500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -525 for b, and -4500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-525\right)±\sqrt{\left(-525\right)^{2}-4\times 15\left(-4500\right)}}{2\times 15}$$

Square -525.

$$x=\frac{-\left(-525\right)±\sqrt{275625-4\times 15\left(-4500\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-525\right)±\sqrt{275625-60\left(-4500\right)}}{2\times 15}$$

Multiply -60 times -4500.

$$x=\frac{-\left(-525\right)±\sqrt{275625+270000}}{2\times 15}$$

Add 275625 to 270000.

$$x=\frac{-\left(-525\right)±\sqrt{545625}}{2\times 15}$$

Take the square root of 545625.

$$x=\frac{-\left(-525\right)±75\sqrt{97}}{2\times 15}$$

The opposite of -525 is 525.

$$x=\frac{525±75\sqrt{97}}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{525±75\sqrt{97}}{30}$$

Now solve the equation x=\frac{525±75\sqrt{97}}{30} when ± is plus. Add 525 to 75\sqrt{97}.

$$x=\frac{75\sqrt{97}+525}{30}$$

Divide 525+75\sqrt{97} by 30.

$$x=\frac{5\sqrt{97}+35}{2}$$

Now solve the equation x=\frac{525±75\sqrt{97}}{30} when ± is minus. Subtract 75\sqrt{97} from 525.

$$x=\frac{525-75\sqrt{97}}{30}$$

Divide 525-75\sqrt{97} by 30.

$$x=\frac{35-5\sqrt{97}}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{97}+35}{2} x=\frac{35-5\sqrt{97}}{2}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$15x^{2}-525x-4500=0$$

Add 4500 to both sides of the equation.

$$15x^{2}-525x-4500-\left(-4500\right)=-\left(-4500\right)$$

Subtracting -4500 from itself leaves 0.

$$15x^{2}-525x=-\left(-4500\right)$$

Subtract -4500 from 0.

$$15x^{2}-525x=4500$$

Divide both sides by 15.

$$\frac{15x^{2}-525x}{15}=\frac{4500}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\left(-\frac{525}{15}\right)x=\frac{4500}{15}$$

Divide -525 by 15.

$$x^{2}-35x=\frac{4500}{15}$$

Divide 4500 by 15.

$$x^{2}-35x=300$$

Divide -35, the coefficient of the x term, by 2 to get -\frac{35}{2}. Then add the square of -\frac{35}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-35x+\left(-\frac{35}{2}\right)^{2}=300+\left(-\frac{35}{2}\right)^{2}$$

Square -\frac{35}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-35x+\frac{1225}{4}=300+\frac{1225}{4}$$

Add 300 to \frac{1225}{4}.

$$x^{2}-35x+\frac{1225}{4}=\frac{2425}{4}$$

Factor x^{2}-35x+\frac{1225}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{35}{2}\right)^{2}=\frac{2425}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{2}\right)^{2}}=\sqrt{\frac{2425}{4}}$$

Simplify.

$$x-\frac{35}{2}=\frac{5\sqrt{97}}{2} x-\frac{35}{2}=-\frac{5\sqrt{97}}{2}$$

Add \frac{35}{2} to both sides of the equation.

$$x=\frac{5\sqrt{97}+35}{2} x=\frac{35-5\sqrt{97}}{2}$$

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 15

$$x ^ 2 -35x -300 = 0$$

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

$$r + s = 35 rs = -300$$

Two numbers r and s sum up to 35 exactly when the average of the two numbers is \frac{1}{2}*35 = \frac{35}{2}. 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>

$$r = \frac{35}{2} - u s = \frac{35}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -300

$$(\frac{35}{2} - u) (\frac{35}{2} + u) = -300$$

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

$$\frac{1225}{4} - u^2 = -300$$

Simplify the expression by subtracting \frac{1225}{4} on both sides

$$-u^2 = -300-\frac{1225}{4} = -\frac{2425}{4}$$

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

$$u^2 = \frac{2425}{4} u = \pm\sqrt{\frac{2425}{4}} = \pm \frac{\sqrt{2425}}{2}$$

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

$$r =\frac{35}{2} - \frac{\sqrt{2425}}{2} = -7.122 s = \frac{35}{2} + \frac{\sqrt{2425}}{2} = 42.122$$