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ग) x^(2)+3x+21=22

ग)\newlinex2+3x+21=22 x^{2}+3 x+21=22

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Full solution

Q. ग)\newlinex2+3x+21=22 x^{2}+3 x+21=22
  1. Subtract 2222: First, let's subtract 2222 from both sides to set the equation to zero.\newlinex2+3x+21−22=0x^2 + 3x + 21 - 22 = 0\newlinex2+3x−1=0x^2 + 3x - 1 = 0
  2. Factor or Use Quadratic Formula: Now, we'll try to factor the quadratic, but it doesn't factor nicely. So, we'll use the quadratic formula instead.\newlineThe quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=3b = 3, and c=−1c = -1.
  3. Plug Values into Formula: Plug the values into the quadratic formula.\newlinex=−3±32−4⋅1⋅(−1)2⋅1x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\newlinex=−3±9+42x = \frac{-3 \pm \sqrt{9 + 4}}{2}\newlinex=−3±132x = \frac{-3 \pm \sqrt{13}}{2}
  4. Find Solutions: So, the solutions are: \newlinex=−3+132x = \frac{-3 + \sqrt{13}}{2} and x=−3−132x = \frac{-3 - \sqrt{13}}{2}

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