What are the solutions to x^2+32x+248=0?(1 point)x=−32+42–√ and x=−32−42–√x=16+22–√ and x=16−22–√x=32+42–√ and

Question

What are the solutions to x^2+32x+248=0?(1 point)x=−32+42–√ and x=−32−42–√x=16+22–√ and x=16−22–√x=32+42–√ and x=32−42–√x=−16+22 −−√ and x=−16−22–√

GPT-5 mini · Answer

Use the quadratic formula:
x = [-32 ± sqrt(32^2 − 4·1·248)]/2 = [-32 ± sqrt(1024 − 992)]/2 = [-32 ± sqrt(32)]/2 = -16 ± 2√2.

So the solutions are x = -16 + 2√2 and x = -16 − 2√2.

What are the solutions to x^2+32x+248=0?(1 point)x=−32+42–√ and x=−32−42–√x=16+22–√ and x=16−22–√x=32+42–√ and x=32−42–√x=−16+22 −−√ and x=−16−22–√