Question

Find of quadratic formula 5 x square - 6 x minus 2 is equal to zero

Answer 1

Answer is in the photo check!!!!!

Answer 2

$\bf\underline{Solution:-}$

Given :-

$\sf\hookrightarrow 5x{}^{2}-6x-2=0$

To find :-

The roots of this quadratic equation

$\bf\underline{\red{\:\:\:\:\:A.T.Q\:\:\:\:}}$

Now by quadratic formula

$\boxed{\sf{x=\dfrac{-b\pm\sqrt{b{}^{2}-4ac}}{2a}}}$

$\sf\implies Here\: 5x{}^{2}-6x-2=0 \:\: \\ \\ \sf\bullet a=5\:\:\:\bullet b=-6\:\:\: \bullet c=-2\\ \\ \sf\implies Put\:this\:values\:in\:the\:formula \\ \\ \sf\implies x=\dfrac{-(-6) \pm\sqrt{(-6){}^{2}-4\times 5\times -2}}{2\times 5}\\ \\ \sf\implies x=\dfrac{6 \pm\sqrt{36-(-40)}}{10}\\ \\ \sf\implies x=\dfrac{6 \pm\sqrt{36+40}}{10}\\ \\ \sf\implies x=\dfrac{6\pm\sqrt{76}}{10}\\ \\ \sf\implies x=\dfrac{6 \pm 2\sqrt{19}}{10}\\ \\ \bf\:\:\:\: Now\:the\: Roots\\ \\ \sf\bullet x=\dfrac{\cancel2(3+\sqrt{19})}{\cancel{10}}\\ \\ \sf\implies x=\dfrac{3+\sqrt{19}}{5}\\ \\ \sf\bullet x=\dfrac{\cancel2(3-\sqrt{19})}{\cancel{10}}\\ \\ \sf\implies x=\dfrac{3-\sqrt{19}}{5}$

$\boxed{\sf{So\:the\:roots=\dfrac{3+\sqrt{19}}{5}\:\:or\:\:\dfrac{3-\sqrt{19}}{5}}}$