SOLUTION: Solve. Round any approximate solution to the fourth decimal place. 20=-3+4(2)^2 Please show me step by step to get the answer. Thanks

SOLUTION: Solve. Round any approximate solution to the fourth decimal place. 20=-3+4(2)^2 Please show me step by step to get the answer. Thanks

Algebra.Com

Question 475936: Solve. Round any approximate solution to the fourth decimal place.
20=-3+4(2)^2
Please show me step by step to get the answer. Thanks
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
No variables. And, 20 is not equal to -3 + 4(2)^2.
RELATED QUESTIONS
Solve. Round any approximate solution to the fourth decimal place. 3^2r x 3^r-4=97 (answered by Tatiana_Stebko)

Solve. Round any approximate solution to the fourth decimal place. 5^x+7=50-3(5)^x (answered by lwsshak3)

Solve: 12t-36=t^2 Please show me step by step to get the answer.... (answered by stanbon)

Simplify: 3^-2 Please show me step by step to get the answer.... (answered by stanbon)

Simplify: -6(bc^4)^-3 Please show me step by step to get the answer.... (answered by stanbon)

factor: the polynomial 2r^2+5r+4 Please show me step by step to get the answer.... (answered by Alan3354)

Simplify: (8b^-2c^2 / 12b^-5c^-3)^-3 Please show me step by step to get the answer.... (answered by Alan3354)

(-7+√69)/2,(-7-√69)/2 These are my exact solutions, need to find... (answered by vleith,richwmiller)

Simplify: (2bc^-7 / 5b^-1c^-2)^3 Please show me step by step to get the answer.... (answered by Alan3354)