Question

$34x ^{2} + 121x + 242 = 0$
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Answer 1

Answer:

$x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a}$

here a = 34 , b = 121 , C = 242

$\frac{ - 121 \binom{ + }{ - } \sqrt{ {121}^{2} - 4 \times 34 \times 242} }{2 \times 34}$

$\frac{ - 121 \binom{ + }{ - } \sqrt{ - 18271} }{68}$

$= \frac{ - 121 \binom{ + }{ - }135.17 }{68} (approximately)$

$hope \: it \:helps..$

THNKS!!

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore x = \frac{ - 121 \pm \sqrt{ - 18271} }{ 68}}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a quadratic eqn.

• We have to find the values of x.

$\underline \bold{Given : } \\ \implies 34 {x}^{2} + 121x + 242 = 0 \\ \\ \underline \bold{To \: Find : } \\ \implies x = ?$

• According to given question :

$\bold{Using \: Quadratic \: formula : } \\ \implies {34x}^{2} + 121x + 242 = 0 \\ \\ \implies D = {b}^{2} - 4ac \\ \\ \implies D = ({121})^{2} - 4 \times 34 \times 242 \\ \\ \implies D= 14641 - 32912 \\ \\ \bold{\implies D = - 18271} \\ \\ \implies x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \implies x = \frac{ - 121 \pm \sqrt{ - 18271} }{2 \times 34} \\ \\ \bold{\implies x = \frac{ - 121 \pm \sqrt{ - 18271} }{ 68} }$