Basic Math Examples

Solve for y -5y(1-5y)+5(-8y-2)=-4y-8y
-5y(1-5y)+5(-8y-2)=-4y-8y
Step 1
Simplify -5y(1-5y)+5(-8y-2).
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Step 1.1
Simplify each term.
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Step 1.1.1
Apply the distributive property.
-5y1-5y(-5y)+5(-8y-2)=-4y-8y
Step 1.1.2
Multiply -5 by 1.
-5y-5y(-5y)+5(-8y-2)=-4y-8y
Step 1.1.3
Rewrite using the commutative property of multiplication.
-5y-5-5yy+5(-8y-2)=-4y-8y
Step 1.1.4
Simplify each term.
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Step 1.1.4.1
Multiply y by y by adding the exponents.
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Step 1.1.4.1.1
Move y.
-5y-5-5(yy)+5(-8y-2)=-4y-8y
Step 1.1.4.1.2
Multiply y by y.
-5y-5-5y2+5(-8y-2)=-4y-8y
-5y-5-5y2+5(-8y-2)=-4y-8y
Step 1.1.4.2
Multiply -5 by -5.
-5y+25y2+5(-8y-2)=-4y-8y
-5y+25y2+5(-8y-2)=-4y-8y
Step 1.1.5
Apply the distributive property.
-5y+25y2+5(-8y)+5-2=-4y-8y
Step 1.1.6
Multiply -8 by 5.
-5y+25y2-40y+5-2=-4y-8y
Step 1.1.7
Multiply 5 by -2.
-5y+25y2-40y-10=-4y-8y
-5y+25y2-40y-10=-4y-8y
Step 1.2
Subtract 40y from -5y.
25y2-45y-10=-4y-8y
25y2-45y-10=-4y-8y
Step 2
Subtract 8y from -4y.
25y2-45y-10=-12y
Step 3
Move all terms containing y to the left side of the equation.
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Step 3.1
Add 12y to both sides of the equation.
25y2-45y-10+12y=0
Step 3.2
Add -45y and 12y.
25y2-33y-10=0
25y2-33y-10=0
Step 4
Use the quadratic formula to find the solutions.
-b±b2-4(ac)2a
Step 5
Substitute the values a=25, b=-33, and c=-10 into the quadratic formula and solve for y.
33±(-33)2-4(25-10)225
Step 6
Simplify.
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Step 6.1
Simplify the numerator.
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Step 6.1.1
Raise -33 to the power of 2.
y=33±1089-425-10225
Step 6.1.2
Multiply -425-10.
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Step 6.1.2.1
Multiply -4 by 25.
y=33±1089-100-10225
Step 6.1.2.2
Multiply -100 by -10.
y=33±1089+1000225
y=33±1089+1000225
Step 6.1.3
Add 1089 and 1000.
y=33±2089225
y=33±2089225
Step 6.2
Multiply 2 by 25.
y=33±208950
y=33±208950
Step 7
The final answer is the combination of both solutions.
y=33+208950,33-208950
Step 8
The result can be shown in multiple forms.
Exact Form:
y=33+208950,33-208950
Decimal Form:
y=1.57411159,-0.25411159
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