Solve (23-10y/4)^2+y^2-(23-10y)-3y-1=0

$$( \frac { 23 - 10 y } { 4 } ) ^ { 2 } + y ^ { 2 } - ( 23 - 10 y ) - 3 y - 1 = 0$$

Solve for y

$$y=\frac{1}{2}=0.5$$

$$y = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using the Quadratic Formula

To raise \frac{23-10y}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-\left(23-10y\right)-3y-1=0$$

To find the opposite of 23-10y, find the opposite of each term.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-23+10y-3y-1=0$$

Combine 10y and -3y to get 7y.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-23+7y-1=0$$

Subtract 1 from -23 to get -24.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-24+7y=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y^{2}-24+7y times \frac{4^{2}}{4^{2}}.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+\frac{\left(y^{2}-24+7y\right)\times 4^{2}}{4^{2}}=0$$

Since \frac{\left(23-10y\right)^{2}}{4^{2}} and \frac{\left(y^{2}-24+7y\right)\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(23-10y\right)^{2}+\left(y^{2}-24+7y\right)\times 4^{2}}{4^{2}}=0$$

Do the multiplications in \left(23-10y\right)^{2}+\left(y^{2}-24+7y\right)\times 4^{2}.

$$\frac{529-460y+100y^{2}+16y^{2}-384+112y}{4^{2}}=0$$

Combine like terms in 529-460y+100y^{2}+16y^{2}-384+112y.

$$\frac{145-348y+116y^{2}}{4^{2}}=0$$

Calculate 4 to the power of 2 and get 16.

$$\frac{145-348y+116y^{2}}{16}=0$$

Divide each term of 145-348y+116y^{2} by 16 to get \frac{145}{16}-\frac{87}{4}y+\frac{29}{4}y^{2}.

$$\frac{145}{16}-\frac{87}{4}y+\frac{29}{4}y^{2}=0$$

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.

$$\frac{29}{4}y^{2}-\frac{87}{4}y+\frac{145}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{29}{4} for a, -\frac{87}{4} for b, and \frac{145}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{87}{4}\right)±\sqrt{\left(-\frac{87}{4}\right)^{2}-4\times \frac{29}{4}\times \frac{145}{16}}}{2\times \frac{29}{4}}$$

Square -\frac{87}{4} by squaring both the numerator and the denominator of the fraction.

$$y=\frac{-\left(-\frac{87}{4}\right)±\sqrt{\frac{7569}{16}-4\times \frac{29}{4}\times \frac{145}{16}}}{2\times \frac{29}{4}}$$

Multiply -4 times \frac{29}{4}.

$$y=\frac{-\left(-\frac{87}{4}\right)±\sqrt{\frac{7569}{16}-29\times \frac{145}{16}}}{2\times \frac{29}{4}}$$

Multiply -29 times \frac{145}{16}.

$$y=\frac{-\left(-\frac{87}{4}\right)±\sqrt{\frac{7569-4205}{16}}}{2\times \frac{29}{4}}$$

Add \frac{7569}{16} to -\frac{4205}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{87}{4}\right)±\sqrt{\frac{841}{4}}}{2\times \frac{29}{4}}$$

Take the square root of \frac{841}{4}.

$$y=\frac{-\left(-\frac{87}{4}\right)±\frac{29}{2}}{2\times \frac{29}{4}}$$

The opposite of -\frac{87}{4} is \frac{87}{4}.

$$y=\frac{\frac{87}{4}±\frac{29}{2}}{2\times \frac{29}{4}}$$

Multiply 2 times \frac{29}{4}.

$$y=\frac{\frac{87}{4}±\frac{29}{2}}{\frac{29}{2}}$$

Now solve the equation y=\frac{\frac{87}{4}±\frac{29}{2}}{\frac{29}{2}} when ± is plus. Add \frac{87}{4} to \frac{29}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{145}{4}}{\frac{29}{2}}$$

Divide \frac{145}{4} by \frac{29}{2} by multiplying \frac{145}{4} by the reciprocal of \frac{29}{2}.

$$y=\frac{5}{2}$$

Now solve the equation y=\frac{\frac{87}{4}±\frac{29}{2}}{\frac{29}{2}} when ± is minus. Subtract \frac{29}{2} from \frac{87}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{29}{4}}{\frac{29}{2}}$$

Divide \frac{29}{4} by \frac{29}{2} by multiplying \frac{29}{4} by the reciprocal of \frac{29}{2}.

$$y=\frac{1}{2}$$

The equation is now solved.

$$y=\frac{5}{2} y=\frac{1}{2}$$

Steps for Completing the Square

To raise \frac{23-10y}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-\left(23-10y\right)-3y-1=0$$

To find the opposite of 23-10y, find the opposite of each term.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-23+10y-3y-1=0$$

Combine 10y and -3y to get 7y.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-23+7y-1=0$$

Subtract 1 from -23 to get -24.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+y^{2}-24+7y=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y^{2}-24+7y times \frac{4^{2}}{4^{2}}.

$$\frac{\left(23-10y\right)^{2}}{4^{2}}+\frac{\left(y^{2}-24+7y\right)\times 4^{2}}{4^{2}}=0$$

Since \frac{\left(23-10y\right)^{2}}{4^{2}} and \frac{\left(y^{2}-24+7y\right)\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(23-10y\right)^{2}+\left(y^{2}-24+7y\right)\times 4^{2}}{4^{2}}=0$$

Do the multiplications in \left(23-10y\right)^{2}+\left(y^{2}-24+7y\right)\times 4^{2}.

$$\frac{529-460y+100y^{2}+16y^{2}-384+112y}{4^{2}}=0$$

Combine like terms in 529-460y+100y^{2}+16y^{2}-384+112y.

$$\frac{145-348y+116y^{2}}{4^{2}}=0$$

Calculate 4 to the power of 2 and get 16.

$$\frac{145-348y+116y^{2}}{16}=0$$

Divide each term of 145-348y+116y^{2} by 16 to get \frac{145}{16}-\frac{87}{4}y+\frac{29}{4}y^{2}.

$$\frac{145}{16}-\frac{87}{4}y+\frac{29}{4}y^{2}=0$$

Subtract \frac{145}{16} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{87}{4}y+\frac{29}{4}y^{2}=-\frac{145}{16}$$

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.

$$\frac{29}{4}y^{2}-\frac{87}{4}y=-\frac{145}{16}$$

Divide both sides of the equation by \frac{29}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{29}{4}y^{2}-\frac{87}{4}y}{\frac{29}{4}}=-\frac{\frac{145}{16}}{\frac{29}{4}}$$

Dividing by \frac{29}{4} undoes the multiplication by \frac{29}{4}.

$$y^{2}+\left(-\frac{\frac{87}{4}}{\frac{29}{4}}\right)y=-\frac{\frac{145}{16}}{\frac{29}{4}}$$

Divide -\frac{87}{4} by \frac{29}{4} by multiplying -\frac{87}{4} by the reciprocal of \frac{29}{4}.

$$y^{2}-3y=-\frac{\frac{145}{16}}{\frac{29}{4}}$$

Divide -\frac{145}{16} by \frac{29}{4} by multiplying -\frac{145}{16} by the reciprocal of \frac{29}{4}.

$$y^{2}-3y=-\frac{5}{4}$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=-\frac{5}{4}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=\frac{-5+9}{4}$$

Add -\frac{5}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-3y+\frac{9}{4}=1$$

Factor y^{2}-3y+\frac{9}{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}.

$$\left(y-\frac{3}{2}\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{1}$$

Simplify.

$$y-\frac{3}{2}=1 y-\frac{3}{2}=-1$$

Add \frac{3}{2} to both sides of the equation.

$$y=\frac{5}{2} y=\frac{1}{2}$$