# Solve the quadratic: -4x^2 – 15x + 4 = 0

Solve the quadratic: $-4{x}^{2}–15x+4=0$
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Willie
The quadratic formula used for quadratic equation $a{x}^{2}+bx+c=0$ is,
$x=-\frac{b±\sqrt{{b}^{2}-4ac}}{2a}$
Here, a=-4, b=-15 ,c=4 therefore
${x}_{1,2}=\frac{-\left(-15\right)±\sqrt{{\left(-15\right)}^{2}-4\left(-4\right)4}}{2\left(-4\right)}$
${x}_{1,2}=\left(15±\frac{\sqrt{289}}{-8}$
${x}_{1,2}=\frac{15±17}{-8}$
${x}_{1}=\frac{15+17}{-8}$
${x}_{1}=\frac{32}{-8}$
${x}_{1}=-4$
and ${x}_{2}=\frac{15-17}{-8}$
${x}_{2}=-\frac{2}{-8}$
${x}_{2}=\frac{1}{4}$
Thus, the solutions of quadratic equation are -4 and $\frac{1}{4}$