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Q: How do you solve this a squared minus b squared plus ac minus bc?

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(b-c)(a+b)-ac

6(b - ac + b2 - bc)

6(ab - ac + b2 - bc)

carry the six then use the quadratic formula negative b plus or minus the squrare root of b squared minus 4 ac all over 2

carry the six then use the quadratic formula negative b plus or minus the squrare root of b squared minus 4 ac all over 2

it means like ac angle or a squared like in the phythangren theorem its a squared + b squared = c squared

negative b plus or minus the square root of b minus 4 times ac all over 2 a

The most surefire way to find the zeroes of a quadratic are to apply the quadratic formula. The formula says that the zeroes of quadratic equations which are generally written as ax2+bx+c=y can be found by taking (-b+/-(b2-4ac).5)/2a or if this notation makes no sense... negative b plus or minus the square-root of b squared minus four ac all over two a. Note: if b squared minus four ac is less than zero, the function has non-real roots

this is what you would: 4x+6=3x-1 -6 -6 ---- 4x=3x-7 -3x -3x ---- 1x=-7 ---- x=-7

Well. Multiply both sides by 'c'. After that you should have something like ac=b-4 then you add 4, to get this result 4+ac=b.

The square root of 149 (7 squared plus 10 squared equals 149).

y=mx+b A=xy; A=(pi)r^2; V=xyh; (-b= or - (b^2-4ac)^1/2)/2a ( actually read as negative b plus or minus the square root of b squared minus four ac, all divided by 2a....quadratic formula) This should get you, if you need more or this is not what you were looking for, let me know.

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