Find the roots of the following quadratic equ | vedclass

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(B) Comparing the equation $y^{2}+10y+6=0$ with the standard form $ay^{2}+by+c=0$,we get $a=1, b=10, c=6$.The discriminant $D$ is given by $D = b^{2}-

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$-5+\sqrt{19}$ and $-5-\sqrt{19}$

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(B) Comparing the equation $y^{2}+10y+6=0$ with the standard form $ay^{2}+by+c=0$,we get $a=1, b=10, c=6$.The discriminant $D$ is given by $D = b^{2}-4ac$.$D = (10)^{2} - 4(1)(6) = 100 - 24 = 76$.Since $D > 0$,the equation has two distinct real roots.The quadratic formula is $y = \frac{-b \pm \sqrt{D}}{2a}$.Substituting the values,we get $y = \frac{-10 \pm \sqrt{76}}{2(1)}$.Since $\sqrt{76} = \sqrt{4 	imes 19} = 2\sqrt{19}$,we have $y = \frac{-10 \pm 2\sqrt{19}}{2}$.Dividing by $2$,we get $y = -5 \pm \sqrt{19}$.Thus,the roots are $-5+\sqrt{19}$ and $-5-\sqrt{19}$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $y^{2}+10y+6=0$

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