Solve 10y^2-35y+30=

$$10 y ^ { 2 } - 35 y + 30 =$$

Factor

$$5\left(y-2\right)\left(2y-3\right)$$

Solution Steps

Factor out 5.

$$5\left(2y^{2}-7y+6\right)$$

Consider 2y^{2}-7y+6. Factor the expression by grouping. First, the expression needs to be rewritten as 2y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 6=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 2y^{2}-7y+6 as \left(2y^{2}-4y\right)+\left(-3y+6\right).

$$\left(2y^{2}-4y\right)+\left(-3y+6\right)$$

Factor out 2y in the first and -3 in the second group.

$$2y\left(y-2\right)-3\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(2y-3\right)$$

Rewrite the complete factored expression.

$$5\left(y-2\right)\left(2y-3\right)$$

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.

$$10y^{2}-35y+30=0$$

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.

$$y=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 10\times 30}}{2\times 10}$$

Square -35.

$$y=\frac{-\left(-35\right)±\sqrt{1225-4\times 10\times 30}}{2\times 10}$$

Multiply -4 times 10.

$$y=\frac{-\left(-35\right)±\sqrt{1225-40\times 30}}{2\times 10}$$

Multiply -40 times 30.

$$y=\frac{-\left(-35\right)±\sqrt{1225-1200}}{2\times 10}$$

Add 1225 to -1200.

$$y=\frac{-\left(-35\right)±\sqrt{25}}{2\times 10}$$

Take the square root of 25.

$$y=\frac{-\left(-35\right)±5}{2\times 10}$$

The opposite of -35 is 35.

$$y=\frac{35±5}{2\times 10}$$

Multiply 2 times 10.

$$y=\frac{35±5}{20}$$

Now solve the equation y=\frac{35±5}{20} when ± is plus. Add 35 to 5.

$$y=\frac{40}{20}$$

Divide 40 by 20.

$$y=2$$

Now solve the equation y=\frac{35±5}{20} when ± is minus. Subtract 5 from 35.

$$y=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

$$y=\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and \frac{3}{2} for x_{2}.

$$10y^{2}-35y+30=10\left(y-2\right)\left(y-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10y^{2}-35y+30=10\left(y-2\right)\times \frac{2y-3}{2}$$

Cancel out 2, the greatest common factor in 10 and 2.

$$10y^{2}-35y+30=5\left(y-2\right)\left(2y-3\right)$$

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

$$x ^ 2 -\frac{7}{2}x +3 = 0$$

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

$$r + s = \frac{7}{2} rs = 3$$

Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{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>

$$r = \frac{7}{4} - u s = \frac{7}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 3

$$(\frac{7}{4} - u) (\frac{7}{4} + u) = 3$$

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

$$\frac{49}{16} - u^2 = 3$$

Simplify the expression by subtracting \frac{49}{16} on both sides

$$-u^2 = 3-\frac{49}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =\frac{7}{4} - \frac{1}{4} = 1.500 s = \frac{7}{4} + \frac{1}{4} = 2$$

Evaluate

$$5\left(y-2\right)\left(2y-3\right)$$