B squared plus 20b equals -80?
Use quadratic equation. I will go thru this one step at a time.
b^2 + 20b = -80 ..add +80 to both sides
b^2 + 20b +80 = 0
General quadratic equation
ax^2 +bx + c = 0
x = [-b ± (b^2 - 4*a*c)^0.5] ÷ 2*a
In your equation
a =1
b = 20
c = 80
Substituting these values into the general equation:
x = [-20 ± (20^2 - 4*1*80)^0.5] ÷ 2*1
x = [-20 ± (400 - 320)^0.5] ÷ 2
x = [-20 ± 80^0.5] ÷ 2
X1 = [-20 + 80^0.5] ÷ 2
X2 = [-20 - 80^0.5] ÷ 2
I got X1 = -5.528 and X2 = -14.47
Or go to a quadratic equation solver web site such as
http://www.math.com/students/calculators/source/quadratic.htm
94+94+80+80=348.
5² + 80² = 25 + 6400 = 6425
80 80 + ββ 160 160 80 + ββ 240
80 + 8 = 88
[ y = ± sqrt(5) ] is.
94+94+80+80=348.
5² + 80² = 25 + 6400 = 6425
80 80 + ββ 160 160 80 + ββ 240
80 + 50 = 130
80 + 8 = 88
[ y = ± sqrt(5) ] is.
The answer is 560. Simply add the numbers together like you would for two plus two. To check your work, a calculator is the best option.
Solve this problem -x squared -40x- 80 =0
245
31,880
190
200