Question

$2x {}^{2} + 3x - 10 = x {}^{2} + 6x + 30$
solve by quadratic formula please? ​

Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Solve:

$:\longrightarrow 2x^{2} + 3x - 10 = x^{2} + 6x + 30 \\\\:\implies 2x^2 + 3x - 10 - (x^2 + 6x + 30) = 0 \\\\:\implies 2x^2 + 3x - 10 - x^2 - 6x - 30 = 0 \\\\:\implies x^2 - 3x - 40 = 0 \\\\:\implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where, a = coefficient of x^2 = 1, b = coefficient of x = -3, and c = constant = -40

$\\:\implies x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-40)}}{2(1)} \\\\:\implies x = \dfrac{3 \pm \sqrt{9 + 160}}{2} \\\\:\implies x = \dfrac{3 \pm \sqrt{169}}{2} \\\\:\implies x = \dfrac{3 \pm 13}{2} \\\\:\implies x = \dfrac{3 + 13}{2} \:\:\:OR\:\:\: x = \dfrac{3 - 13}{2} \\\\:\implies x = \dfrac{16}{2} \:\:\:OR\:\:\: x = \dfrac{-10}{2} \\\\\bf{:\implies x = 8 \:\:\:OR\:\:\: x = -5}$

Hope it helps!!!!