Solve 30x^2-360x+1050=0

$$30 x ^ { 2 } - 360 x + 1050 = 0$$

Solve for x

$$x=5$$

$$x=7$$

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.

$$30x^{2}-360x+1050=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 30 for a, -360 for b, and 1050 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-360\right)±\sqrt{\left(-360\right)^{2}-4\times 30\times 1050}}{2\times 30}$$

Square -360.

$$x=\frac{-\left(-360\right)±\sqrt{129600-4\times 30\times 1050}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-\left(-360\right)±\sqrt{129600-120\times 1050}}{2\times 30}$$

Multiply -120 times 1050.

$$x=\frac{-\left(-360\right)±\sqrt{129600-126000}}{2\times 30}$$

Add 129600 to -126000.

$$x=\frac{-\left(-360\right)±\sqrt{3600}}{2\times 30}$$

Take the square root of 3600.

$$x=\frac{-\left(-360\right)±60}{2\times 30}$$

The opposite of -360 is 360.

$$x=\frac{360±60}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{360±60}{60}$$

Now solve the equation x=\frac{360±60}{60} when ± is plus. Add 360 to 60.

$$x=\frac{420}{60}$$

Divide 420 by 60.

$$x=7$$

Now solve the equation x=\frac{360±60}{60} when ± is minus. Subtract 60 from 360.

$$x=\frac{300}{60}$$

Divide 300 by 60.

$$x=5$$

The equation is now solved.

$$x=7 x=5$$

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.

$$30x^{2}-360x+1050=0$$

Subtract 1050 from both sides of the equation.

$$30x^{2}-360x+1050-1050=-1050$$

Subtracting 1050 from itself leaves 0.

$$30x^{2}-360x=-1050$$

Divide both sides by 30.

$$\frac{30x^{2}-360x}{30}=-\frac{1050}{30}$$

Dividing by 30 undoes the multiplication by 30.

$$x^{2}+\left(-\frac{360}{30}\right)x=-\frac{1050}{30}$$

Divide -360 by 30.

$$x^{2}-12x=-\frac{1050}{30}$$

Divide -1050 by 30.

$$x^{2}-12x=-35$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-35+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-35+36$$

Add -35 to 36.

$$x^{2}-12x+36=1$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-6=1 x-6=-1$$

Add 6 to both sides of the equation.

$$x=7 x=5$$

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 30

$$x ^ 2 -12x +35 = 0$$

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

$$r + s = 12 rs = 35$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 6. 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>

$$r = 6 - u s = 6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 35

$$(6 - u) (6 + u) = 35$$

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

$$36 - u^2 = 35$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 35-36 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =6 - 1 = 5 s = 6 + 1 = 7$$