SOLUTION: please can you explain how to find the solution to the quadratic equation with brackets in x(2x+5)=10

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Question 215290This question is from textbook Higher GCSE Mathematics for Edexcel
: please can you explain how to find the solution to the quadratic equation with brackets in
x(2x+5)=10 This question is from textbook Higher GCSE Mathematics for Edexcel

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x%282x%2B5%29=10 Start with the given equation.


2x%5E2%2B5x=10 Distribute


2x%5E2%2B5x-10=0 Subtract 10 from both sides.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=2, b=5, and c=-10


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%285%29+%2B-+sqrt%28+%285%29%5E2-4%282%29%28-10%29+%29%29%2F%282%282%29%29 Plug in a=2, b=5, and c=-10


x+=+%28-5+%2B-+sqrt%28+25-4%282%29%28-10%29+%29%29%2F%282%282%29%29 Square 5 to get 25.


x+=+%28-5+%2B-+sqrt%28+25--80+%29%29%2F%282%282%29%29 Multiply 4%282%29%28-10%29 to get -80


x+=+%28-5+%2B-+sqrt%28+25%2B80+%29%29%2F%282%282%29%29 Rewrite sqrt%2825--80%29 as sqrt%2825%2B80%29


x+=+%28-5+%2B-+sqrt%28+105+%29%29%2F%282%282%29%29 Add 25 to 80 to get 105


x+=+%28-5+%2B-+sqrt%28+105+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


x+=+%28-5%2Bsqrt%28105%29%29%2F%284%29 or x+=+%28-5-sqrt%28105%29%29%2F%284%29 Break up the expression.


So the answers are x+=+%28-5%2Bsqrt%28105%29%29%2F%284%29 or x+=+%28-5-sqrt%28105%29%29%2F%284%29


which approximate to x=1.312 or x=-3.812