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Question Solve the quadratic equation 2y(y+5)−5=02y(y+5)-5=0 using the quadratic formula. Provide the solution(s) in the form y=□y=\square.

Studdy Solution

STEP 1

Assumptions
1. The equation given is 2y(y+5)−5=02y(y+5)-5=0.
2. We need to solve for yy using the Quadratic Formula.

STEP 2

First, we need to expand the equation and write it in the standard quadratic form, ay2+by+c=0ay^2 + by + c = 0.
2y(y+5)−5=2y2+10y−52y(y+5)-5=2y^2+10y-5

STEP 3

Now, we can identify the coefficients aa, bb, and cc from the quadratic equation.
a=2,b=10,c=−5a = 2, \quad b = 10, \quad c = -5

STEP 4

The Quadratic Formula is given by:
y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

STEP 5

Substitute the values of aa, bb, and cc into the Quadratic Formula.
y=−10±102−4(2)(−5)2(2)y = \frac{-10 \pm \sqrt{10^2 - 4(2)(-5)}}{2(2)}

STEP 6

Simplify the expression under the square root (the discriminant).
102−4(2)(−5)=100+40=140\sqrt{10^2 - 4(2)(-5)} = \sqrt{100 + 40} = \sqrt{140}

STEP 7

Further simplify the square root by factoring out perfect squares.
140=4⋅35=235\sqrt{140} = \sqrt{4 \cdot 35} = 2\sqrt{35}

STEP 8

Now, substitute the simplified discriminant back into the Quadratic Formula.
y=−10±2354y = \frac{-10 \pm 2\sqrt{35}}{4}

STEP 9

Simplify the fraction by dividing both the numerator and the denominator by the common factor of 2.
y=−5±352y = \frac{-5 \pm \sqrt{35}}{2}

STEP 10

We have two solutions for yy, one with the plus sign and one with the minus sign.
y1=−5+352,y2=−5−352y_1 = \frac{-5 + \sqrt{35}}{2}, \quad y_2 = \frac{-5 - \sqrt{35}}{2}
The solutions to the equation 2y(y+5)−5=02y(y+5)-5=0 are:
y=−5+352,−5−352 y=\frac{-5 + \sqrt{35}}{2}, \frac{-5 - \sqrt{35}}{2}

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