Question

Pls help only if yk it ill give brainliest to the correct answer !

Answer 1

Answer:

$x=-0.3$

$x=-11.7$

Step-by-step explanation:

$4x^2+24x+13=2x^2+6\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\4x^2+24x+13-6=2x^2+6-6\\Simplify\\4x^2+24x+7=2x^2\\\mathrm{Subtract\:}2x^2\mathrm{\:from\:both\:sides}\\4x^2+24x+7-2x^2=2x^2-2x^2\\Simplify\\2x^2+24x+7=0\\\mathrm{Solve\:with\:the\:quadratic\:formula}\\Quadratic\:Equation\:Formula\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=24,\:c=7:\quad x_{1,\:2}=\frac{-24\pm \sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}\\x=\frac{-24+\sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}:\quad \frac{-12+\sqrt{130}}{2}\\x=\frac{-24-\sqrt{24^2-4\cdot \:2\cdot \:7}}{2\cdot \:2}:\quad -\frac{12+\sqrt{130}}{2}\\The\:solutions\:to\:the\:quadratic\:equation\:are:\\x=\frac{-12+\sqrt{130}}{2},\:x=-\frac{12+\sqrt{130}}{2}\\\\quad x=-0.29912\ ,\:x=-11.70087$