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4x^2+x-5=-6x
We move all terms to the left:
4x^2+x-5-(-6x)=0
We get rid of parentheses
4x^2+x+6x-5=0
We add all the numbers together, and all the variables
4x^2+7x-5=0
a = 4; b = 7; c = -5;
Δ = b2-4ac
Δ = 72-4·4·(-5)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{129}}{2*4}=\frac{-7-\sqrt{129}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{129}}{2*4}=\frac{-7+\sqrt{129}}{8} $
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