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Precalculus Examples
x2+30x+200=0x2+30x+200=0
Step 1
Use the quadratic formula to find the solutions.
-b±√b2-4(ac)2a−b±√b2−4(ac)2a
Step 2
Substitute the values a=1a=1, b=30b=30, and c=200c=200 into the quadratic formula and solve for xx.
-30±√302-4⋅(1⋅200)2⋅1−30±√302−4⋅(1⋅200)2⋅1
Step 3
Step 3.1
Simplify the numerator.
Step 3.1.1
Raise 3030 to the power of 22.
x=-30±√900-4⋅1⋅2002⋅1x=−30±√900−4⋅1⋅2002⋅1
Step 3.1.2
Multiply -4⋅1⋅200−4⋅1⋅200.
Step 3.1.2.1
Multiply -4−4 by 11.
x=-30±√900-4⋅2002⋅1x=−30±√900−4⋅2002⋅1
Step 3.1.2.2
Multiply -4−4 by 200200.
x=-30±√900-8002⋅1x=−30±√900−8002⋅1
x=-30±√900-8002⋅1x=−30±√900−8002⋅1
Step 3.1.3
Subtract 800800 from 900900.
x=-30±√1002⋅1x=−30±√1002⋅1
Step 3.1.4
Rewrite 100100 as 102102.
x=-30±√1022⋅1x=−30±√1022⋅1
Step 3.1.5
Pull terms out from under the radical, assuming positive real numbers.
x=-30±102⋅1x=−30±102⋅1
x=-30±102⋅1x=−30±102⋅1
Step 3.2
Multiply 22 by 11.
x=-30±102x=−30±102
Step 3.3
Simplify -30±102−30±102.
x=-15±5x=−15±5
x=-15±5x=−15±5
Step 4
The final answer is the combination of both solutions.
x=-10,-20x=−10,−20